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PLANE  GEOMETRY 


WILLIS 


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PREFACE 

This  text  represents  the  experience  and  developing  ideals 
of  a  quarter  of  a  century  of  geometry  teaching.  The 
philosophy  and  the  methods  set  forth  in  it  are  those  of  the 
author's  own  class-room.  He  believes  that  a  pupil  learns 
more  of  the  subject  when  it  is  presented  as  in  these  pages, 
and  he  is  certain  that  every  pupil  who  is  taught  by  this 
method  under  the  guidance  of  a  teacher  who  is  himself 
interested  in  his  subject,  becomes  inspired  to  learn  and  is 
willing  to  work  to  learn.  An  awakening  of  a  love  of  the 
subject  is  a  service  that  may  be  rendered  to  our  pupils 
of  every  intellectual  grade.  The  text  is  so  arranged  that  it 
may  be  variously  used. 

A  SHORT  COURSE. — From  the  complete  material  pre- 
sented in  this  text,  what  may  be  considered  as  the  funda- 
mentals, or  as  constituting  a  minimum  requirement,  has 
been  designated  by  a  BLACK  FACE  paragraph  numeral. 
It  constitutes  a  Short  Course. 

As  much  as  possible  of  the  remaining  material  should 
be  included.  This  remaining  material,  which  completes  and 
rounds  out  the  subject,  is  designated  by  a  LIGHT  FACE 
paragraph  numeral. 

A  LABORATORY  COURSE. — Chapter  1  and  the  chapters 
which  contain  the  Experimental  Determination  of  Principles 
may  be  used  as  a  laboratory  course,  omitting  all  of  the 
chapters  containing  the  Classification  and  Explanation  of 
Principles  except  the  most  elementary  portions. 

A  COMPLETE  COURSE. — The  text  is  primarily  intended  to 
be  used  as  a  whole,  without  many  omissions.  It  contains 
all  that  need  be  required  of  any  class.  The  experimental 
chapters  should  be  rigorously  required  of  all,  and  the 
accurate  or  free-hand  construction  of  all  figures  should  be 
insisted  upon  throughout  the  portions  devoted  to  classifica- 
tion and  explanation  of  principles. 

541281 


VI  PREFACE 

EXERCISES. — A  sufficient  number  of  exercises,  carefully 
selected  and  graded,  will  be  found  in  the  list  of  Additional 
Theorems,  Exercises  and  Review  Exercises  to  enable  the 
teacher  to  vary  his  class  assignments  from  term  to  term  for 
many  years.  No  class  should  be  expected  to  solve  all. 

FOR  EXCEPTIONAL  PUPILS. — The  wealth  of  material 
presented  affords  a  splendid  means  of  developing  the  pupil 
of  exceptional  ability.  Extra  work  for  additional  credit 
may  be  allowed  every  pupil.  It  is  better  to  stimulate  a 
pupil  to  greater  interest  and  accomplishment,  especially,  in  a 
subject  so  vast  and  offering  so  great  rewards  as  geometry, 
than  to  hurry  an  able  pupil  through  the  subject,  almost 
always  to  his  loss  rather  than  to  his  profit. 

DEDICATION. — The  conception  of  this  work  runs  far  back 
in  the  years.  It  is  the  author's  contribution  in  the  service 
of  his  profession.  It  is  dedicated  to  all  those  many  students 
of  the  grand  old  science,  who  perhaps  unknowingly,  and  to 
all  those  teachers  who  have  keenly  felt  the  need  of  a  more 
sympathetic  interpretation  of  its  immortal  truths. 

In  the  hope  that  it  will  influence  better  teaching  and 
learning,  and  will  help  to  upbuild  and  uphold  a  generation 
which  will  reverence  our  beloved  geometry  for  its  own  sake, 
as  well  as  for  its  uses  inseparable  from  a  great  constructive 
civilization — the  author  sends  this  little  volume  forth  upon 
its  mission. 

C.  ADDISON  WILLIS. 

GIRARD  COLLEGE,  PHILADELPHIA. 


CONTENTS 

CHAPTER  I 

PAGE 

FIRST  PRINCIPLES 1 

CHAPTER  II 

STRAIGHT  LINES,  CIRCLES,  SYMMETRY 13 

Principles  Determined  Experimentally. 

CHAPTER  III 

PARALLELS,  PERPENDICULARS,  ANGLES 30 

Classification  and  Explanation  of  Principles. 

CHAPTER  IV 

TRIANGLES 

Principles  Determined  Experimentally. 

CHAPTER  V 

TRIANGLES 63 

Classification  and  Explanation  of  Principles. 

CHAPTER  VI 

POLYGONS 90 

Principles  Determined  Experimentally. 

CHAPTER  VII 

POLYGONS 97 

Classification  and  Explanation  of  Principles. 

CHAPTER  VIII 

Loci 116 

Principles  Determined  Experimentally. 

CHAPTER  IX 

Loci 123 

Classification  and  Explanation  of  Principles. 

CHAPTER  X 

THE  MEASUREMENT  OF  SECTS,  SIMILAR  TRIANGLES 129 

Principles  Determined  Experimentally. 

vii 


Vlll  CONTENTS 

CHAPTER  XI 

PAGE 

SIMILAR  TRIANGLES,  SIMILAR  POLYGONS 143 

Classification  and  Explanation  of  Principles. 

CHAPTER  XII 

CIRCLES 178 

Principles  Determined  Experimentally. 

CHAPTER  XIII 

CIRCLES 194 

Classification  and  Explanation  of  Principles. 

CHAPTER  XIV 

REGULAR  POLYGONS,  CIRCLES  ASSOCIATED  WITH  POLYGONS.    .    .   236 
Principles  Determined  Experimentally. 

CHAPTER  XV 

REGULAR  POLYGONS  AND  CIRCLES 243 

Classification  and  Explanation  of  Principles. 

CHAPTER  XVI 

AREAS. 260 

Principles  Determined  Experimentally. 

CHAPTER  XVII 

AREAS 269 

Classification  and  Explanation  of  Principles. 

INDEX  .  .   299 


CHAPTER  I 
FIRST  PRINCIPLES 

1.  Note-book. — Preserve  all  drawings  and  written  work 
in  a  loose  leaf  note-book  with  unruled  pages.     Number  to 
correspond  with  the  text.     Draw  large,  accurate  figures 
with  distinct  lettering. 

2.  Drawing  Instruments. — Each   pupil  should  be  pro- 
vided with  one  six-inch  triangle  with  30°,  60°  and  90° 
angles,  a  six-inch  scale   divided   decimally,   a  protractor, 
and  a  compass. 

3.  How  Geometry  Began. — Many  ancient  nations,  the 
Chinese,   Babylonians,   Egyptians,   Greeks,   Romans,   and 
others,    constructed    wonderful   works,    as   roads,    canals, 
aqueducts,    temples,    pyramids,    and   tombs.     They  laid 
out  cities  and  gardens.     These  required  appreciation  of 
geometric  principles.     Herodotus,  a  Greek  historian  who 
wrote  about  450  B.C., -states  that  the  Egyptians  developed 
many  geometric  principles  because  the  annual  overflow  of 
the  Nile  obliterated  the  boundary  marks  of  the  lands,  and 
it  became  necessary  to  re-locate  these  boundaries.     This 
was  probably  the  earliest  use  of  surveying  for  land  measure- 
ment, and  it  is  from  this  use  of  geometric  principles  that 
the  name  geometry,  which  means  "earth  measurement," 
is  derived. 

EXERCISES 

1.  What  ru:er  first  appointed  surveyors  in  Egypt?     Refer  to  an 
article  on  geometry  in  an  encyclopedia. 

2.  Look  up  the  derivation  of  the  word  geometry.     Name  some  other 
familiar  English  words  containing  one  or  the  other  of  these 
Greek  root-words. 

4.  The  Modern  Value  of  Geometry. — Geometry  forms  the 
structure    of    many    modern    sciences    and    arts.     Artists 

1 


PLANE    GEOMETRY 

apply  its  principles  in  pictures  and  designs;  architects  use 
them  in  drawing  plans;  civil  engineers  and  surveyors  use 
geometry  now  as  they  did  centuries  ago;  astronomers  could 
not  measure  nor  describe  the  motions  and  positions  of  the 
stars  or  planets  without  it;  nor  could  navigators  find  their 
place  at  sea.  Many  artisans,  also,  and  workers  in  machine 
shops,  carpenters,  stone  cutters,  and  others,  make  constant 
use  of  geometric  principles. 

5.  Familiar    Use    of   Geometric    Terms. — Every    word 
which  has  to  do  with  position,  form  or  size,  is  geometric. 

EXERCISES 

1.  How  far  is  a  point  on  the  surface  of  the  earth  from  the  center 
of  the  earth?     How  tall  are  you? 

2.  What  is  the  form  of  the  earth?     Of  a  piece  of  crayon? 

3.  What  is  the  area  of  the  floor  of  the  class-room  in  square  feet? 

4.  What  is  the  volume  of  the  class-room  in  cubic  feet? 

5.  What  kind  of  surface  are  the  floor  and  walls? 

6.  What  is  the  form  of  the  uncut  end  of  a  lead  pencil? 

7.  What  is  the  form  of  a  base-ball  diamond? 

8.  What  is  the  form  of  a  tennis  court? 

9.  What  is  the  form  of  a  Rugby  football? 

10.  Write  such  formulas  of  mensuration  (geometric  measurement) 
as  you  can  recall.  Have  these  formulas  practical  use?  Many 
of  them  are  derived  and  explained  in  this  text. 

6.  The  Subject  Matter. — Geometry  considers  the  prop- 
erties of  solids,   surfaces,  lines  and  points.     A  solid  is  a 
definite  portion  of  space.     Thus  the  properties  of  the  space, 
or  room,  occupied  by  an  object  belong  to  geometry,  and 
not  the  physical  properties  of  the  object. 

A  surface  is  the  boundary  of  a  solid. 

A  line  is  the  boundary  of  a  surface. 

A  point  is  the  end  of  a  line,  or  divides  a  line. 

7.  Dimensions. — A  dimension  is  extent  of  some  definite 
kind. 

A  point  has  no  dimensions;  that  is,  it  has  no  extent.     It 
has  position  only. 

A  line  has  one  dimension,  length. 


FIRST   PRINCIPLES  O 

A  surface  has  two  dimensions,  length  and  width  (or 
breadth). 

A  solid  has  three  dimensions,  length,  width,  and  height. 

EXERCISES 

1.  Refer  to  a  point  in  the  room;  where  is  it? 

2.  Refer  to  a  line;  how  long  is  it? 

3.  Refer  to  a  surface;  how  long  and  how  wide  is  it? 

4.  Refer  to  a  solid;  estimate  its  three  dimensions. 

8.  Points. — A  point  is  named  by  a  capital  letter. 

EXERCISES 

1.  Draw  a  point  •  or  X.     Name  it. 

2.  Draw  a  straight  line.     Mark  a  point  on  it  1 


9.  Lines. — A  straight  line  is  the  shortest  line  that  can 
be  drawn,  or  imagined,  between  two  points. 

A  curved  line  is  one  no  part  of  which  is  straight. 

A  sect  is  a  straight  line  of  definite  length. 

A  unit  sect  is  a  sect  of  generally  known  or  standard 
length,  as  the  inch,  meter,  mile. 

A  sect  is  measured  in  terms  of  some  unit  sect.  The 
measurement  is  a  number  which  expresses  how  many  times 
the  given  sect  contains  the  unit  sect. 

A  line  is  named  by  two  capital  letters,  or  by  a  single 
letter. 

A         B_  t        x      [     y     [ 

A  sect  is  bisected  when  it  is  divided  by  a  point  or  inter- 
secting line  into  two  equal  parts. 

EXERCISES 

1.  From  what  word  is  straight  derived? 

2.  What  instrument  is  used  to  draw  a  straight  line? 

3.  What  instrument  contains  a  unit  sect? 

4.  What  instruments  may  be  used  to  copy  a  given  sect? 

5.  Draw  a  sect  of  3  inches;  one  of  3.4  inches. 

6.  Draw  a  sect  as  long  as  the    ^                        *.                   B 
sect  AB,  or  x.  -* ^— 


4  PLANE    GEOMETRY 

7.  Measure  the  sect  AB. 

8.  What  drawing  instrument  is  used  to  draw  a  curved  line? 
What  kind  of  curved  line  may  be  drawn  with  it? 

9.  May  this  curved  line  be  measured  in  the  same  units  as  a 
straight  line? 

10.  Draw  such  a  curved  line;  measure  its  length  in  inches. 

11.  Bisect  (by  judgment)  each  of  the  sects  of  Example  5  by  a  point. 

12.  Bisect  the  sect  of  Example  6  by  a  line. 

13.  Draw  and  trisect  a  sect. 

14.  Can  an  indefinite  line  be  bisected? 

15.  Draw  and  define  a  broken  line. 

16.  Draw  and  define  a  mixed  line. 

10.  Positive   and   Negative   Sects. — These   terms   refer 
to  the  direction  in  which  a  sect  is  drawn  or  measured.     In 
general,    sects   are    considered   positive.     Sometimes  it  is- 
necessary,   or  convenient,   to  distinguish  between  a  sect 
drawn,  or  measured,  from  left  to  right  and  one  drawn,  or 
measured,  from  right  to  left.     The  former  direction  is  then 
generally  considered  positive,  and  the  latter  negative. 

EXERCISES 

1.  Draw  a  sect  and  read  it  as  a  positive  sect.     Read  it  as  a  negative 
sect. 

2.  If  the  distance  and  direction  of  a  point  A  from  a  point  B  are 
indicated  on  a  surveyor's  plot  by  +  785.46  feet;  how  are  the  dis- 
tance and  direction  of  point  B  from  point  A  indicated? 

3.  Read  sect  AC  as  the  sum     A  B  C 

of  two  sects.    Read  AB  as  H ' ' 

the   (algebraic)  sum  of  two  sects,  one  positive  and  the  other 
negative. 

4.  A  man  walks  from  point  A,  —250  feet  to  point  B;  and  from  point 
B,  +300  feet  to  point  C.     Make  a  figure  to  show  where  point  C 
is  located  with  reference  to  point  A. 

11.  Planes. — A  plane,  or  a  plane  surface,  is  such  a  surface 
that  a  straight  line  between  any  two  points  of  the  surface 
lies  wholly  within  it. 

Plane  geometry  is  the  geometry  of  figures  which  are,  or 
could  be,  drawn  on  a  plane. 

EXERCISES 

1.  How  does  a  carpenter  test  the  planeness  of  a  board?     Does  the 
method  agree  with  the  definition  of  a  plane? 


FIRST   PRINCIPLES  5 

2.  Why  is  a  carpenter's  plane  so  called?     Why  is  a  certain  machine 
tool  called  a  planer? 

3.  Define  a  plane  curve.     Refer  to,  or  form  with  a  string  or  wire, 
a  curve  that  is  not  plane. 

4.  Are  two  pencils  with  points  in  contact,  in  the  same  plane? 

5.  What  kind  of  surfaces  are  cylindrical  and  conical  surfaces? 

12.  Straight  Lines  in  Relative  Position. — Two  straight 
lines  in  the  same  plane  may  be  intersecting,  parallel  or 
coinciding.     A  system  of  parallels  consists  of  three  or  more 
parallel  lines. 

Three  or  more  lines  are  concurrent  which  have  a  common 
point  of  intersection.     They  form  a  pencil  of  lines. 

EXERCISES 

1.  Draw  two  parallels;  a  system  of  parallels;  two  intersecting  lines; 
four  concurrent  lines;  two  coinciding  lines;  a  pencil  of  three  lines. 

2.  Define  parallel  lines. 

3.  Hold  two  pencils  so  that  they  are  not  intersecting,  parallel  or 
coinciding.     Are  they  in  the  same  plane?     Refer  to  two  such 
lines  in  the  room. 

13.  Generation  by  Motion. — The  line,  surface  and  solid 
may  be  regarded  as  generated  by  motion. 

EXERCISES 

1.  When  a  point  moves,  what  does  it  generate;  that  is,  what  does 
the  path  of  the  point  represent? 

2.  If  a  line  moves,  what  does  it  generate? 

3.  If  a  surface  moves,  what  does  it  generate? 

14.  Circles. — A  circle  is  a  portion  of  a  plane  bounded  by 
a  curved  line  every  point  of  which  is  equidistant  from  a 
fixed  point  within  it,  called  the  center. 

The  circumference  is  the  bounding  line; 
which  is  also  called  a  circle. 

A  radius  is  a  straight  line  joining  the 
center  with  a  point  on  the  circum- 
ference. 

A  chord  is  a  straight  line  joining  two 
points  of  the  circumference. 

A  diameter  is  a  chord  which  passes  through  the  center. 


6  PLANE   GEOMETRY 

A  tangent  is  a  straight  line  which  touches  the  circumfer- 
ence in  only  one  point  however  far  it  is  extended. 
An  arc  is  a  portion  of  a  circumference. 
A  unit  arc  is  an  arc  of  standard  length. 

A  degree  of  arc  (1°)  is         of  a  circumference. 


The  degree  is  the  unit  arc  which  is  most  used. 

The  measurement  of  an  arc  is  a  number  which  expresses 
how  many  times  the  given  arc  contains  the  unit  arc  of  the 
same  circle. 

A  minute  of  arc  is  ™  of  a  degree. 

A  second  of  arc  is  ^  of  a  minute. 

A  semicircle  is  one-half  of  a  circle. 

A  quadrant  is  one-quarter  of  a  circle,  or  of  a  circumference. 
Every  circle  contains  360°  of  arc,  but  the  size  of  a  circle 
is  given  by  the  length  of  its  radius  or  diameter. 
A  circle  is  named  by  a  capital  letter  at  the  center. 

EXERCISES 

1.  Draw  a  circle.     Draw  a  radius,  diameter,  chord,  tangent,  degree 
of  arc,  quadrant,  semicircle,  semicircumference. 

2.  Draw  a  curve  which  is  not  a  circle. 

3.  What  instrument  is  used  to  measure  arcs  of  circles? 

4.  How  many  degrees  of  arc  are  there  in  a  quadrant?     In  a  semi- 
circumference?     How  many  quadrants  are  in  a  semicircumfer- 
ence or  semicircle? 

5.  Draw  a  circle  and  mark  arcs  of  approximately  60°;  120°;  45°; 
30°;  1°. 

6.  On  a  circle,  lay  off  an  arc  AB  =  +160°;  from  point  B,  lay  off 
an  arc  BC  =  —100°.     Read  the  value  of  arc  CA. 

15.  Relative  Position  of  Two  Circles. — Two  circles  of 
the  same,  or  different,  radii  may  be  situated  in  relation  to 
each  other  in  three  ways:  (1)  out  of  contact;  (2)  tangent;  (3) 
intersecting. 

Concentric  circles  have  the  same  center  and  different 
radii. 


FIRST   PRINCIPLES  7 

EXERCISES 

1.  Draw  two  circles  out  of  contact  in  two  different  ways  (internally 
and  externally  out  of  contact). 

2.  Draw  two  circles  tangent  in  two  different  ways  (internally  and 
externally). 

3.  Draw  two  intersecting  circles. 

4.  Draw  two  concentric  circles;  three  concentric  circles. 

5.  What  class  includes  concentric  circles? 

16.  Angles. — An  angle  is  a  figure  formed  by  two  straight 
lines  drawn  from  the  same  point. 

G 


The  parts  of  an  angle  are  the  sides  and  the  vertex. 

An  angle  may  be  considered  as  having  been  generated  by 
the  rotation  of  one  of  its  sides  about  the  vertex  from  a  posi- 
tion of  coincidence  with  the  other  side.  An  angle  may  be 
positive  or  negative  according  to  the  direction  of  this  rota- 
tion. The  counter-clockwise  direction  is  generally  regarded 
as  positive. 

A  measurement  arc  is  often  used  to  indicate  the  size  or 
direction,  or  both,  of  an  angle. 

An  angle  is  named  (1)  by  three  capital  letters,  reading 
from  one  side  to  the  other  with  the  vertex  letter  in  the 
middle ;  (2)  when  no  confusion  can  occur,  by  the  vertex  let- 
ter alone;  (3)  by  a  small  letter,  or  number,  or  capital  letter 
placed  within  the  angle. 

EXERCISES 

1.  Read  the  angles  of  Figures  1  and  2  in  all  possible  ways. 

2.  Define  the  vertex  and  sides  of  an  angle. 

3.  Draw  two  lines,  A  B  and  CD,  intersecting  at  point  0.     Read  all 
the  angles  which  are  formed. 


8  PLANE    GEOMETRY 

17.  Angles  Named  According  to  Size. 

A  zero  angle  is  one  whose  sides  coincide  before 
rotation. 

A  straight  angle  is  one  whose  sides  lie  in  the  same  straight 
line   in   opposite   directions   from   the 
vertex. 

A  right  angle  is  one-half  of  a  straight  angle. 

A  perigon  is  an  angle  whose  sides  coincide  ^\ 

after  rotation. 

An  acute  angle  is  less  than  a  right  angle. 

An  obtuse  angle  is  greater  than  a  right  angle  and  less  than 
a  straight  angle. 

A  reflex  angle  is  greater  than  a  straight  angle  and  generally 
is  not  more  than  a  perigon. 

An  oblique  angle  is  any  angle  other  than  a  multiple  of  a 
right  angle. 

Two  angles  whose  sum  is  a 
right  angle  are  complementary. 

Two  angles  whose  sum  is  a  straight  angle  are  supplement- 
ary. 

Two  angles  whose  sum  is  a  perigon  are  explementary. 

The  angles  are  complements,  supplements,  explements  of 
each  other. 

A  unit  angle  is  an  angle  of  standard  size. 

A  degree  of  angle  is  §V  of  a  right  angle. 

The  measurement  of  an  angle  is  a  number  which  expresses 
how  many  times  the  given  angle  contains  the  unit  angle. 

EXERCISES 

1.  Draw  angles  of  each  kind  named  above,  with  their  measurement 
arcs. 

2.  Calculate  the  complement  and  supplement  of  72°. 

3.  What  angle  has  a  negative  complement  and  a  positive  supple- 
ment? 

4.  Estimate  the  angles  x,  y  and  DEG  of  paragraph  16;  in  right 
angles,  degrees  and  straight  angles. 

5.  Draw  angles  with  the  protractor  to   the  following  measures: 
1  degree,  or  1°;  45°;  130°;  2£  right  angles;  180°;  If  straight  angles. 


FIRST    PRINCIPLES 


9 


6.  What  angle  occurs  most  frequently  in  familiar  objects? 

7.  Is  the  size  of  an  angle  determined  by  the  length  or  by  the  posi- 
tion of  its  sides? 

18.  Angles  Named  According  to  Relative  Position. — 
Unless  otherwise  stated,  these  names  refer  only  to  two 
angles. 

Adjacent  angles  have  a  common  side  and  vertex. 

Opposite  angles  (or  vertical  angles)  are  the  non-adjacent 
angles  formed  by  two  intersecting  lines. 

Supplementary-adjacent  angles  are  both  supplementary 
and  adjacent. 

A  transversal  is  a  line  which  intersects  two  or  more 
other  lines. 

The  names  given  to  pairs  of  angles,  one  at  each  point  of 
intersection,  are: 


Corresponding;  r  and  w. 
Alternate-interior;  u  and 


x. 


y. 


Alternate-exterior;  rand 


Consecutive-interior;  I 
and  x. 

Consecutive-exterior;  s 
and  y. 


D 


EXERCISES 

1.  Draw  two  externally  adjacent  angles.     Draw  two  internally 
adjacent  angles.     Draw  four  successively  adjacent  angles. 

2.  Draw  two  intersecting  lines.     Name  all  pairs  of  opposite  angles. 

3.  Draw  two  supplementary-adjacent  angles.     Draw  two  supple- 
mentary angles  that  are  not  adjacent. 

4.  Name  all  other  pairs  of  corresponding  angles  formed  in  the  trans- 
versal   figure;  of    alternate-interior  angles;  of  alternate-exterior 
angles;   of  consecutive-interior  angles;  of  consecutive-exterior 
angles. 

5.  Draw  three  lines  cut  by  a  transversal.     Name  a  set  of  corres- 
ponding angles. 


10  PLANE   GEOMETRY 

19.  Other  Angle  Names. — A  perpendicular  is  a  straight 
line  which  forms  a  right  angle  with  another  straight  line. 
The  foot  of  the  perpendicular  is  the  point  of  intersection. 

A  bisector  of  an  angle  is  a  straight  line  which  divides  it 
into  two  equal  angles. 

EXERCISES 

1.  Draw  AB  perpendicular  to  CD  at  point  A.     Name  the  foot  of 
the  perpendicular  A  B.     Is  line  CD  also  perpendicular  to  line  A  B  ? 

2.  Can  an  angle  bisector  be  drawn  otherwise  than  through  the 
vertex? 

3.  Draw  acute,  right,  obtuse,  reflex  and  straight  angles,  and  a 
perigon.     Draw  by  judgment  the  bisector  of  each. 

4.  Draw  and  trisect  an  angle? 

20.  Distance. — The  distance  between  two  points  is  the 
length  of  the  straight  line  joining  the  points. 

The  distance  from  a  point  to  a  straight  line  is  the  length 
of  a  perpendicular  drawn  from  the  point  to  the  line. 

The  distance  between  two  parallels  is  measured  on  a 
perpendicular  drawn  to  both  parallels. 

The  distance  from  a  point  to  a  circle  is  measured  on  a 
radius  of  the  circle,  or  on  a  radius  prolonged,  which  is  drawn 
through  the  given  point. 

EXERCISES 

1.  Mark  two  points.     Draw  and  measure  the  distance  between 
them. 

2.  Mark  a  point  and  draw  a  straight  line.     Draw  and  measure  the 
distance  from  the  point  to  the  line. 

3.  Draw  two  parallels.     Draw  and  measure  the  distance  between 
them. 

4.  Draw  a  circle  and  mark  a  point  within  it.     Draw  and  measure 
the  distance  from  the  point  to  the  circumference. 

5.  Mark  a  point  outside  the  circle  which  is  two  inches  from  the 
circumference. 

21.  REVIEW  EXERCISES 

1.  Give  the  geometric  names  of  various  objects  in  the  class-room. 

2.  What  kinds  of  angles  are  formed  by  the  corners  of  your  draw- 
ing triangles?     Measure  them  in  degrees. 


FIBST   PRINCIPLES  11 

3.  Name  two  streets  that  intersect  at  a  right  angle;  two  that  inter- 
sect at  an  oblique  angle;  two  streets  that  will  not  intersect  no 
matter  how  far  they  are  extended  in  either  direction. 

4.  Can  a  sect  be  measured  in  degrees?     Or  an  angle  in  inches? 
Has  a  degree  of  angle  any  equivalent  in  inches? 

5.  Can  an  arc  be  measured  in  inches? 

6.  If  you  walk  from  a  given  point  A,  +250  feet  to  a  point  B;  and 
then  walk  from  point  B,    —  400  feet  to  a   point  C;  how  is 
point  C  situated  with  respect  to  point  A?     Make  a    drawing 

7.  If  the  terminal  side  of  an  angle  rotates  from  the  initial  side  A  B, 
+  120  degrees  to  the  position  AC;  and  then  rotates  from  the 
position  AC,  —200   degrees  to  the  position  AD;  how  is  the 
terminal  side  AD  situated  with  respect  to  the  initial  side  AB? 

8.  Judge  the  measure  of  each  of  these  angles  in  terms  of  the  right 
angle;  also  in  terms  of  the  degree. 


At. 


9.  Reduce  f  of  a  right  angle  to  degrees,  minutes  and  seconds. 

10.  Reduce  22°  30'  (22  degrees  and  30  minutes)  to  a  fraction  of  a 
right  angle. 

11.  How  many  degrees  of  angle  has  the  minute  hand  of  a  clock 
described  (generated)  from  twelve  o'clock  to  one  minute  (of 
time)  past  twelve? 

12.  How  many  degrees  of  angle  are  between  a  line  running  from  a 
certain  point  northeast,   and   a  line  from  the    same  point 
due  east? 

13.  Calculate  and  draw  the  supplement  of  200  degrees;  the  com- 
plement of  100  degrees. 

14.  Write  the  complement  of  x  degrees;  the  supplement.     Write 
the  complement  of  x  right  angles;  the  supplement. 

15.  Find  the  angle  whose  supplement  equals  three  times  the  com- 
plement.    Help.  180°  -  x  =  3(90°  -  »). 

16.  Express  algebraically  the  relation  that  x  and  y  are  comple- 
mentary; supplementary. 

17.  One  of  two  complementary  angles  is  10°  less  than  twice  the 
other  angle.     Find  the  angles. 

18.  The   difference   between   two   supplementary   angles   is    25°. 
Find  the  angles. 

19.  Upon  what  does  sign  depend  in  sects,  angles  and  arcs? 

20.  What  angle-bisector  is  a  perpendicular  to  one  of  the  sides  of 
the  angle? 

21.  Measure    the    distance    from    this  point    to  the  nearest  and 
farthest  edges  of  this  page. 


12 


PLANE    GEOMETRY 


22.  What  line  is  the  distance  from  the  center  of  a  circle  to  the 
circumference? 

23.  On  what  line  is  the  distance  measured  between  two  concentric 
circles? 

24.  On  what  line  is  the  distance  measured  between  the  circumfer- 
ences of  two  circles  that  are  out  of  contact? 

25.  Define  parallels  by  means  of  a  property  of  the  distance  between 
them. 

26.  Show  that  in  every  use  of  the  term  distance  between  two  objects, 
the  shortest  distance  is  referred  to. 

27.  Draw  two  straight  lines  which  do  not  coincide,  which  form  a  zero 
angle  at  infinity. 

22.   ABBREVATIONS  AND   SYMBOLS 


adjacent adj. 

angle,  angles Z,    Zs 

alternate-interior alt. -int. 

alternate-exterior alt.-ext. 

circle,  circles O,  © 

consecutive cons. 

corresponding corres. 

degree,  minute,  second .  ° 

equal,  is  equal  to = 

equals =s 

is  unequal  to 7^ 

unequals F^S 

exterior ext. 

interior. .  .  int. 


feet,  inches '     " 

is  greater  than > 

is  less  than < 

opposite opp. 

parallel,  is  parallel  to ...    1 1 

parallels ||s 

perpendicular,  is  perpen- 
dicular to _L 

perpendiculars _l_s 

point,  points pt,  pts. 

right rt. 

straight st. 

supplementary-adjacent  sup. -adj. 
paragraph  reference. .  .  .    l,26,etc. 


CHAPTER  II 
STRAIGHT  LINES,  CIRCLES,  SYMMETRY 

EXPERIMENTAL  INVESTIGATION  OF  PRINCIPLES 

23.  Geometrical  Experiments. — The  facts  of  geometry 
are  all  about  us.     We  intend  to  inquire  into  nature's  laws 
and  to  discover  these  facts  through  our  own  powers  of 
observation.     For  this  purpose  a  series  of  experiments  will 
be  undertaken;  an  experiment  being  an  investigation  made 
for  the  purpose  of  ascertaining  a  fact. 

We  shall  make  a  drawing  according  to  some  assigned 
(or  given)  property,  and  shall  observe  some  new  property 
which  is  invariably  associated  with  the  given  property. 
The  finding  of  this  new  property  is  the  result  of  the 
experiment. 

Make  all  experiments  for  the  general  case,  so  that  results 
shall  not  be  limited  in  their  application.  State  the  result 
in  the  shortest  and  clearest  expressions  possible.  Intro- 
duce nothing  outside  of  the  experiment.  If  an  experiment 
is  inconclusive,  do  not  hesitate  to  so  state  the  result.  You 
are  not  to  state  what  you  think  ought  to  be  true,  but  exactly 
what  the  experiment  shows  to  be  true.  Sometimes  it  is 
well  to  repeat  an  experiment  one  or  several  times. 

EXERCISES 

1.  Are  other  sciences  based  upon  experiment? 

2.  How  may  the  equality  or  difference  of  two  sects  be  tested  by  the 
use  of  instruments?     Of  two  angles? 

3.  How  may  the  perpendicularity,  or  lack  of  it,  of  two  lines  be 
tested? 

24.  Experiment  I. — The  number  of  straight  lines  that 
can  be  drawn  through  given  points. 

(a)  One  given  point. 

(b)  Two  given  points. 

(c)  Three  or  more  given  points. 

13 


14  PLANE    GEOMETRY 

(a)  Mark  a  point  A.  Draw  a  straight  line  through  it. 
If  possible,  draw  other  straight  lines  through  point  A. 
Observe  the  number  of  straight  lines  which  can  be  drawn 
through  point  A,  and  their  directions. 

RESULT. — Any  number  of  straight  lines  can  be  drawn 
through  a  given  point,  and  in  any  direction. 

(6)  Mark  two  points  A  and  B.  Draw  a  straight  line 
through  (containing)  both  points.  If  possible,  draw  other 
straight  lines  containing  both  points,  in  different  positions 
from  the  line  first  drawn.  Observe  the  number  of  straight 
lines  which  can  be  so  drawn.  State  result. 

(c)  Mark  three  points  at  random.  Observe  the  number 
of  straight  lines  which  can  be  drawn  containing  all  the 
points.  Can  a  straight  line  be  drawn  containing  any  three 
marked  points?  Make  the  experiment  for  four  points. 

RESULT. — In  general,  no  straight  line  can  be  drawn 
through  three  or  more  given  points.  But  the  points  may 
be  so  situated  that  one  straight  line  shall  contain  them. 

25.  Experiment  II. — Prolongation  of  a  straight  line. 
Draw  a   (straight)   line.     Prolong   (extend)   it  in  both 

directions,  using  dotted  lines.  Observe  how  far  it  can  be 
thus  extended  if  the  drawing  is  not  limited  by  the  extent  of 
the  paper. 

RESULT. — 

26.  EXPERIMENT  III. — The  number  of  points  of  intersec- 
tion of  straight  lines  and  circles. 

(a)  Two  straight  lines. 

(b)  One  straight  line  and  one  circle. 

(c)  Two  circles  (of  the  same  or  different  radii) . 

In  stating  the  results,  observe  that  in  certain  positions 
of  the  straight  lines  and  circles  there  may  be  no  points  of 
intersection,  and  in  other  positions  there  may  be  one  or 
more  points  of  intersection  or  of  contact.  State  the  com- 
plete result  for  each  part. 

27.  Constructions. — A  construction  is  a   drawing  made 
for  the  purpose  of  producing  a  figure  with  some  required 


STRAIGHT   LINES,    CIRCLES,    SYMMETRY  15 

property.     The  method  of  obtaining  this  required  figure  is 
the  subject  of  experiment. 

The  different  construction  methods  are: 

(1)  Rule  and  compass. 

(2)  Measurement. 

(3)  Trial  and  error. 

(4)  Copying  a  standard. 

28.  Experiment  IV. — To  draw  (or  construct)  the  bisector 
of  an  angle.  That  is:  An  experiment  to  find  a  method  of 
drawing  the  bisector  of  an  angle. 

C 


/.ABC  is  the  given  angle; 
Required  to  construct  its  bisector. 
METHOD  (a). — (1)  Draw  an  arc,  center  at  B,  any  radius, 
intersecting  BA  and  BC  in  points  D  and  E. 

(2)  Draw  two  arcs,  centers  at  D  and  E,  any  equal  radii 
greater  than  J  DE,  intersecting  each  other  at  F. 

(3)  Draw  BF,  which  is  the  required  bisector  of  /.ABC. 
METHOD  (b). — (1)  Measure   /.ABC  with  the  protractor. 

(2)  Divide  the  measurement  by  2. 

(3)  Lay  out   /.ABG  equal  to  the  quotient. 

EXERCISES 

1.  After  obtaining  the  bisector  BG,  how  may    /sABG  and  GBC 
be  compared  in  order  to  test  their  equality? 

2.  Can  any  angle,  straight,  obtuse  or  reflex,  be  bisected  by  these 
methods? 

3.  Can  another  different  bisector  of   /.ABC  be  drawn? 

4.  What  methods  as  defined  in  27,  have  been  employed  in  the 
preceding  experiment? 

5.  Can  an  angle  be  divided  into  three,  four,  five,  etc.,  equal  angles 

by  either  of  these  methods? 


16  PLANE    GEOMETRY 

29.  Experiment  V.— To  bisect  a  sect. 

Make  the  construction:  (a)  using  the  rule  and  compass 
method;  (6)  using  the  measurement  method;  (c)  using  the 
trial  and  error  method. 

A  CED  C     BD 

-H h-fh 1 M 


AB  is  the  given  sect; 
Required  to  bisect  it. 


METHOD  (c). — (1)  Lay  off  with  the  compass  any  length 
AC  judged  =  JA£,  and  lay  off  CC'  =  AC. 

(2)  If  C'  does  not  coincide  with  B,  set  the  compass  to  a 
different  length  as  AD,  and  make  DD'  =  AD. 

(3)  Repeat  until  the  middle  point  of  A  B  is  found,  as  at  E. 

EXERCISES 

1.  At  how  many  points  can  a  sect  be  bisected? 

2.  Can  a  sect  be  divided  into  any  number  of  equal  parts? 

3.  Which  of  the  methods  of  the  preceding  experiment  can  be  used 
to  divide  a  sect  into  three,  four,  five,  etc.,  equal  parts? 

30.  Experiment  VI. — To  construct  a  perpendicular  to  a 
line  at  a  point  in  the  line. 

Find  (a)  a  method  by  rule  and  compass;  (b)  a  method 
by  measurement;  (c)  a  method  by  copying  a  standard. 

D 


AB  is  the  given  line,  and  C  is  a  point  in  AB; 
Required  to  draw  a  perpendicular  to  A  B  at  point  C. 
METHOD    (c). — (1)  Place   a   standard   right   angle   with 
vertex  at  C,  and  one  side  coinciding  with  CB. 

(2)  Draw  CD,  which  is  the  required  perpendicular. 


STRAIGHT    LINES,    CIRCLES,    SYMMETRY  17 

EXERCISES 

1.  Can  a  perpendicular  be  drawn  (erected)  at  any  point  in  the  line? 

2.  How  many  perpendiculars  can  be  erected  at  a  given  point  as  C, 
on  one  side  of  line  AB1 

3.  Methods  (a)  and  (6)  in  the  preceding  experiment  are  equivalent 
to  bisecting  what  kind  of  angle? 

31.  Experiment  VII. — To  construct  a  perpendicular  to  a 
line  from  a  point  without  (outside  of)  it. 

Draw  and  describe  methods  (a)  by  rule  and  compass; 
(6)  by  copying  a  standard.  . 

32.  Experiment  VIII. — To  construct  an  angle  equal  to  a 
given  angle ;  or,  to  copy  a  given  angle. 

Draw  and  describe  methods  (a)  by  rule  and  compass; 
(b)  by  measurement. 

EXERCISES 

1.  Use   these  methods  to  draw  an  angle 
equal  to  a  given  obtuse  angle. 

2.  Use  the  method  indicated  in  the  figure 
to   copy    a   given    angle.      This    is    a 
modification     of     the     measurement 
method. 

33.  EXPERIMENT  IX. — To  copy  a  sect;  addition  and  sub- 
traction of  sects. 

Draw  two  sects.  Describe  different  methods  of  copying 
them,  of  drawing  a  sect  equal  to  their  sum,  and  of  drawing 
a  sect  equal  to  their  difference. 

34.  EXPERIMENT  X. — Addition  and  subtraction  of  angles. 
Draw  two  angles;  etc. 

35.  EXPERIMENT  XL — To  draw  the  complement  and  the 
supplement  of  a  given  angle. 

Draw  an  angle.  Find  (a)  a  method  by  rule  and  com- 
pass; (6)  a  method  by  measurement;  of  drawing  the  re- 
quired angles. 

36.  Experiment    XII. — The    relative    size    of    opposite 
angles. 

Draw  two  intersecting  lines.  Observe  the  size  of  opposite 
angles.  State  result. 


18  PLANE   GEOMETRY 

37.  Experiment    XIII. — The    relative    position    of    two 
perpendiculars  to  the  same  straight  line. 

Draw  the  figure.     State  result. 

38.  EXBERIMENT  XIV. — To  construct  a  line  through  a 
given  point  parallel  to  a  given  line. 

Draw  the  given  line  and  the  given  point.     Use  the 
principle  of  37  to  complete  the  required  figure. 

EXERCISES 

1.  Can  a  line  be  drawn  through  any  given  point  parallel  to  any 
given  line? 

2.  How  many  lines,  in  different  positions,  can  be  drawn  through 
a  given  point  parallel  to  a  given  line? 

39.  Experiment  XV. — The  position  of  two  lines  cut  by 
a  transversal  making  a  pair  of  corresponding  angles  equal. 

,C 


To  Draw  the  Required  Figure. — (1)   Draw  AB  cut  by 
transversal  CD,  forming   Zz. 

(2)  Mark  point  F  on  CD;  construct    Z.y  =    Z.x,  and  in 
such  a  position  that  it  forms  with   Z.x  a  pair  of  correspond- 
ing angles;  using  either  the  rule  and  compass  method  of 
copying  the  angle,  as  shown,  or  copying  the  angle  with  a 
protractor. 

(3)  Draw  GH,  the  side  of    Z.y. 

Observe  the  position  of  A B  and  GH.     The  result  should 
be  a  complete  statement  of  the  property  of  the  figure,  thus : 


STRAIGHT   LINES,    CIRCLES,    SYMMETRY  19 

RESULT. — Two  straight  lines  which  are  cut  by  a  trans- 
versal so  as  to  make  a  pair  of  corresponding  angles  equal, 
are  parallel. 

EXERCISES 

1.  May  any  pair  of  corresponding  angles  be  made  equal  with  the 
same  result? 

2.  Draw  a  figure  in  which  the  protractor  is  used  to  make  Z.y  =    Z.x. 

40.  Experiment  XVI. — The   position  of  two   lines  cut 
by  a  transversal  making  a  pair  of  alternate -interior  angles 
equal. 

Draw  the  figure.     State  result. 

41.  EXPERIMENT  XVII. — The  position  of  two  lines  cut  by 
a  transversal  making  a  pair  of  alternate -exterior  angles 
equal. 

Draw  the  figure.     State  result. 

42.  Experiment  XVIII. — The  position  of  two  lines  cut 
by  a  transversal  making  a  pair  of  consecutive -interior 
angles  supplementary. 


To  Draw  the  Figure. — (1)  Draw  supplementary  angles 
x  and  y. 

(2)  Draw  AB,  mark  points  C  and  D. 

(3)  Draw  Zs  x'  and  y'  equal  to   Zs  x  and  y  respectively, 
and  EF  and  GHy  the  sides  of    Zs  x'  and  yf.     Or  measure 

Zs  x'  and  y'  with  a  protractor,  so  that  their  sum  equals 
180°.     State  result. 


20  PLANE    GEOMETRY 

43.  EXPERIMENT  XIX. — The  position  of  two  lines  cut  by 
a  transversal  making  a  pair  of  consecutive -exterior  angles 
supplementary. 

Draw  the  figure.     State  result. 

44.  Experiment  XX. — To  construct  a  line  through  a  given 
point  parallel  to  a  given  line. 

This  figure  has  been  constructed  c 

by  means  of  two  perpendiculars  to 
the  same  line;  38.  Employ  in  the 
present  construction  the  principles 
of  39,  40  and  41. 

45.  EXPERIMENT  XXI. — To  construct  a  line  through  a 
given  point  parallel  to  a  given  line,  by  using  rule  and 
triangle. 

This  construction  can  be  made  by  placing  the  rule  and 

triangle  in  certain  positions, 
and  by  then  sliding  the  triangle 
along  the  rule.  Find  the 
best  positions  for  the  rule  and 
triangle.  This  is  an  accurate 
and  rapid  method  of  drawing 
parallels  and  is  the  best 
method  for  general  use.  It  is 
much  used  by  draughtsmen. 

EXERCISES 

1.  What  represents  the  transversal  in  this  construction?     What 
pair  of  angles  is  made  equal? 

2.  To  which  of  the  methods  of  44  is  this  method  equivalent? 

3.  The  rule  may  be  considered  as  a  pair  of  standard  parallels. 
Draw  two  parallel  lines  by  this  instrument  alone.     May  a  line 
be  drawn  through   any  given  point  parallel  to  any  given  line 
by  this  method? 

46.  Experiment  XXII. — Relations  between  pairs  of  angles 
when  two  parallels  are  cut  by  a  transversal. 

Draw  two  parallels  by  any  method. 

(a)  Draw  a  new  transversal.  Observe  the  relative  size  of 
each  pair  of  corresponding,  alternate-interior,  etc.  angles. 


'    STRAIGHT   LINES,    CIRCLES,    SYMMETRY  21 

(6)  Draw  a  perpendicular  to  one  of  the  parallels  and 
observe  its  position  with  respect  to  the  other  parallel. 
State  results  for  both  (a)  and  (6) . 

47.  EXPERIMENT  XXIII. — The  position  of  two  parallels 
to  the  same  line. 

Draw  the  figure.     State  result. 

48.  Experiment  XXIV. — Relation  of  angles  whose  sides 
are  parallel,  each  to  each. 

To  Draw  the  Figure. — (1)  Draw  an  oblique  angle. 
(2)     Draw     other     angles  ^«C^ — 

whose  sides  are  parallel,  each       **£*. ^       c 

GIVEN    ANGLE 

to  each,  to  those  of  the  given  .^ 

angle.     There  are  four  posi- 
tions in  which  such  angles  can  be  drawn,  two  of  these 
positions  being  shown  in  the  figure  at  B  and  C. 

Observe  the  relation  between  the  given  angle  and  the 
constructed  angles.  State  result. 

49.  Experiment  XXV. — Relation  of  angles  whose  sides 
are  perpendicular,  each  to  each. 

Draw  the  figure  and  state  result. 

50.  EXPERIMENT  XXVI. — Properties  of  a  perpendicular. 
(a)  Observe  the  relative  lengths  of  a  perpendicular  and 

an  oblique  line  drawn  from  the  same  given  point  to  the 
same  given  line.  State  result. 

(6)  Draw  a  line  and  a  perpendicular  to  it;  mark  a  point 
in  the  perpendicular;  measure  two  equal  distances  on  the 
given  line  from  the  foot  of  the  perpendicular  (one  on  each 
side  of  it);  draw  two  oblique  lines  from  the  point  in  the 
perpendicular  to  the  ends  of  the  equal  distances. 

Observe  how  the  lengths  of  the  two  oblique  lines  compare. 
State  result. 

(c)  Make  an  experiment  similar  to  part  (6),  in  which  the 
two  distances  measured  on  the  given  line  from  the  foot  of 
the  perpendicular  are  unequal.  State  result. 

SYMMETRY 

51.  THE   SYMMETRY   OF   PLANE   FIGURES   is    OF    Two 
KINDS: — (1)  A  figure  possesses  symmetry  with  respect  to  an 


22 


PLANE    GEOMETRY 


axis,  which  is  a  straight  line,  when  it  can  be  folded  upon 
the  axis  so  that  the  part  of  the  figure  on  one  side  of  the  axis 
coincides  with  the  part  on  the  other  side.  The  axis  is  called 
the  axis  of  symmetry. 

(2)  A  figure  possesses  symmetry  with  respect  to  a  center, 
which  is  a  point,  when,  after  rotation  about  the  center 
through  any  integral  part  of  a  perigon,  it  coincides  with  its 
original  position.  The  point  is  called  the  center  of  symmetry. 

Symmetry  with  respect  to  a  center  is  two-fold,  .three- 
fold, etc.,  according  to  the  integral  division  of  a  perigon 
through  which  it  is  rotated  into  coincidence  with  the 
original  position. 

Any  one  element,  or  symmetrical  division,  of  a  figure 
is  the  symmetrical  impression  of  any  other  element. 

EXERCISES 

1.  Name  the  kind  of  symmetry  possessed  by  each  of  these  figures. 


2.  Draw  a  square.     Draw  all  possible  axes  of  symmetry.     Show 
that  it  possesses  two-fold  and  four-fold  symmetry  with  respect 
to  a  center. 

3.  Go  over  your  note-book  and  select  all  the  figures  possessing 
symmetry  of  any  kind. 

4.  How  many  axes  of  symmetry  has  a  circle?     How  many  kinds 
of  symmetry  with  respect  to  a  center  has  it? 

52.  EXPERIMENT  XXVII. — The  law  of  symmetry  of  a 
point  with  respect  to  an  axis. 

Draw  a  line  AB  and  a  point  C  without  it.  Fold  the 
paper  on  AB  and  trace  the  symmetrical  impression  C",  of 
point  0.  Unfold  the  paper;  draw  CC'. 

Observe  the  relation  of  CC'  to  the   axis  AB. 

State  this  relation  as  the  result  of  the  experiment. 


STRAIGHT   LINES,    CIRCLES,    SYMMETRY  23 

53.  EXPERIMENT  XXVIII. — To  construct  the  symmetrical 
impression  with  respect  to  an  axis : 

(a)  Of  a  sect. 

(b)  Of  an  indefinite  straight  line. 

(c)  Of  an  irregular  curve. 

(a)  Find  the  symmetrical  impression  of  the  two  end 
points  of  the  sect;  join  them  by  a  straight  line. 

(b)  Find  the  symmetrical  impression  of  any  two  points 
of  the  given  line,  etc. 

(c)  Find  the   symmetrical  impression  of  an  indefinite 
number  of  points  of  the  given  curve;  draw  a  curve  con- 
taining them. 

Test  the  accuracy  of  each  construction  by  folding  the 
paper  on  the  axis. 

EXERCISES 

1.  Construct  the  symmetrical  impression  of  a  perpendicular  with 
respect  to  the  line  on  which  it  stands,  as  an  axis. 

2.  Construct  the  symmetrical  impression  of  a  line  parallel  to  the 
axis. 

3.  Construct  the  symmetrical  impression  of  a  line  intersecting  the 
axis.     Show  also  how  this  can  be  done  by  copying  an  angle. 

4.  Construct  the  symmetrical  impression  of  a  circle  with  respect  to 
an  axis  outside  the  circle.     What  is  the  least  number  of  points 
which  will  determine  the  symmetrical  circle? 

54.  EXPERIMENT  XXIX. — The  law  of  symmetry  of  a  point 
with  respect  to  a  center : 

Mark  two  points  A  and  B.  Fix  a  piece  of  transparent 
paper  with  a  pin  at  A ;  trace  point  B  upon  the  transparent 
paper.  Revolve  the  transparent  paper  through  a  straight 
angle;  trace  the  new  position  of  point  B  on  the  drawing  at 
B'.  Remove  the  transparent  paper  and  draw  BB'. 

Observe  the  relation  of  BB'  to  the  center  of  symmetry  A . 
State  result. 

55.  EXPERIMENT  XXX. — The  two-fold  symmetrical  im- 
pression with  respect  to  a  center : 

(a)0fasect. 

\  (b)  Of  an  indefinite  straight  line . 
(c)  Of  an  irregular  curve.  A 


24 


PLANE    GEOMETRY 


Test  the  accuracy  of  each  con- 
struction by  tracing  the  figure 
on  transparent  paper  and  re- 
volving on  the  center  through 
a  straight  angle. 

EXERCISES 

1.  Construct  the  symmetrical  im- 
pression of  a  sect  with  respect  to 
a  center  at  one  of  its  extremities. 

2.  Construct   a  design  with  three- 
fold  symmetry  from  the  given 
element  xyz.     What  must  be  the 

measure  of  angle  a? 

3.  Are  doorways,  windows,  mantlepieces, 
building   outlines,   symmetrical? 

4.  Are  most  pieces  of  furniture  symmetri- 
cal?    Name  some  with  one  or  the  other 
kind  of  symmetry. 

5.  Examine  any  pleasing  outline  of  design,  landscape  gardening, 
architecture,    etc.     Do    all    possess   symmetry   or   a   modified 
symmetry? 

6.  Examine   some   good   group    pictures    or   statuary.     Do   they 
possess  a  certain  balance  of  proportion  and  grouping  that  may 
be  regarded  as  symmetry? 

7.  Are  flowers,  leaves,  entire  plants,  symmetrical? 

56.  EXPERIMENT  XXXI. — Geometrical  illusions. 

The  properties  of  some  geometric  figures  are  always 
judged  incorrectly.  Draw  the  following  figures  much 
enlarged.  It  is  well  to  make  blackboard  drawings  of  some  of 
them. 

(a)  Draw  four  vertical  lines,  equally  spaced. 
Rule  an  oblique  line  as  shown.  Do  the  parts 
of  the  oblique  line  appear  to  lie  in  the  same 
straight  line? 

(6)  Measure  a  perpendicular  equal  in  length  to 
the  horizontal  line.     Do  they  appear  equal? 


STRAIGHT    LINES,    CIRCLES,    SYMMETRY 


25 


(c)  Divide  one  of  two  adjacent  right 
angles  into  a  number  of  equal  parts. 
State  apparent  size  of  the  two  right 
angles. 

(d!)  Draw  two  equal  parallel  lines.  Attach  the 
"arrow-head"  and  " feather"  marks  at  the  ends. 
State  result. 

(e)  Draw  two  horizontal  parallel  lines.     From 
a  point  A  midway  between  them,  draw  radial  lines  /  \ 
as  shown  at  A',  at  about  equal  angles. 
State  result. 

(/)  Draw  four  parallel  lines.  Draw 
sets  of  short  oblique  parallel  lines 
across  them.  State  result. 

GENERAL  RESULTS. — Our  judgment 
of  geometrical  figures  may  lead  us  into 
errors.      The    results  of   the    experi- 
ments performed  in  this  chapter  have  been 
stated  on  the  assumption,  which  is  probably 
a  safe  one,  that  nature  prefers  parallels,  per- 
pendiculars,   equalities,    universal   laws    and 
general  simplicity. 

57.  REVIEW  EXERCISES 

1.  Draw  opposite  angles;  bisect  one  of  them;  prolong  the  bisector 
through  the  other  angle  of  the  pair.     How  does  it  divide  the 
other  angle?     State  result. 

2.  Divide  a  given  sect  into  four  equal  parts. 

3.  Construct  accurate  angles  of  90°  and  45°,  using  rule  and  com- 
pass.    Test  these  angle  values  of  your  protractor  and  triangles. 

4.  Divide  a  given  obtuse  angle  into  four  equal  parts. 

5.  Draw  with  the  protractor  an  angle  of  25°.     Extend  both  sides 
through  the  vertex.     Calculate  and  measure  the  other  three 
angles  thus  formed. 

6.  Draw  a  triangle  ABC  and  a  point  D  without  it.     Construct  the 
two-fold  symmetrical  impression  of  ABC  with  respect  to  the 
center  D, 

7.  Draw  a  circle  with  a  chord  AB.     Construct  the  symmetrical 
impression  of  the  circle  about  the  chord  as  an  axis. 

8.  Which  of  the  printed  Gothic  capitals,  A,  B,  C,  etc.,  possess 


26  PLANE   GEOMETRY 

symmetry  with  respect  to:  (a)  a  vertical  axis;  (6)  a  horizontal 
axis;  (c)  both  vertical  and  horizontal  axes;  (d)  a  center  and  not 
to  any  axes;  (e)  a  center  and  also  one  or  two  axes;  (/)  oblique 
axes?  (g)  Which  letters  are  asymmetrical  (not  symmetrical)  ? 
9.  State  all  the  methods  that  may  be  used  to  draw  a  line  through 
a  given  point  parallel  to  a  given  line.  ~ 

10.  Construct  a  design  from  the  given  element  xy.  B 
Use:    (1)  AB  as  an  axis;  (2)  AC  as  an  axis;  (3)       ^"/V 
DE  as  an  axis.     What  kinds  of  symmetry  will 

the  complete  design  possess?  ^ 

11.  Mark   a  point  on   the  floor.     With  a  piece  of 
string  find  the  shortest  length  from  the  point  to 
the   wall.      Draw   this   position   of   the   string. 
What  line  will  this  position  be? 

12.  Draw  two  sects  AB  and  CD:  (a)  perpendicular  bisectors  of 
each  other,  but  not  equal;  (6)  so  that  AB  is  the  perpendicular 
bisector  of  CD,  but  AB  is  not  bisected;  (c)  so  that  each  bisects 
the  other;  but  not  perpendicular;  (d)  so  that  they  intersect  but 
neither  sect  is  bisected. 

A- 


7 


B- 

13.  Draw  two  parallel  lines  as  shown.     Draw  radial  lines  from 
points  A  and  B,  terminating  in  line  CD.     State  the  illusion 
result. 

14.  When  a  draughtsman  draws  parallels  by  sliding  his  T-square 
along  the  edge  of  his   board,  what  construction  principle  is 
employed? 

APPLICATIONS 

58.  Surveying,  Building  and  Designing. — A  surveyor 
uses  a  tape  to  measure  or  lay  out  lines  (sects),  and  a  transit 
to  measure  or  lay  out  angles.  A  carpenter  or  mason  uses  a 
tape  or  rule  to  measure  or  lay  out  lines. 

The  tape  is  a  flexible  scale,  usually  50  or  100  feet  long. 
The  transit  is  an  accurate  portable  protractor.  The  transit 
consists  of  an  accurately  graduated  horizontal  circle, 
sometimes  also  a  vertical  circle  or  part  of  such  a  circle,  a 


STRAIGHT   LINES,    CIRCLES,    SYMMETRY 


27 


telescope  provided  with  two  intersecting  cross-hairs  at  the 
focal  point  within  the  tube  of  the  telescope,  spirit  levels  and 
levelling  screws  for  levelling  the  graduated  circle,  clamps 
and  slow  motion  screws  for  clamping  or  rotating  the  gradu- 


ated circle  or  the  telescope,  either  separately  or  together, 
and  a  tripod  for  mounting  the  instrument. 

The  transit  is  set  up  over  the  vertex  of  the  angle  which 
is  to  be  measured  or  laid  out.     The  graduated  circle  is 


28  PLANE    GEOMETRY 

levelled  and  the  telescope  focussed  for  cross-hairs  and  dis- 
tance. The  telescope  is  then  set  with  the  " upper  motion" 
so  that  the  index  reading  of  the  circle  is  0°.  While  clamped 
in  this  position,  the  circle  and  telescope  are  rotated  together 
with  the  " lower  motion"  until  the  vertical  cross-hair  of 
the  telescope  cuts  a  point  which  marks  one  side  of  the 
angle  to  be  measured  or  laid  out.  The  upper  clamp  is 
loosened  and  the  telescope  alone  is  rotated  into  the  position 
of  the  other  side  of  the  angle.  The  reading  of  the  index 
on  the  graduated  circle  then  gives  the  value  of  the  angle. 
A  text  on  surveying  should  be  consulted  for  more  complete 
instructions  in  the  use  of  a  transit,  and  for  the  best  methods 
of  measuring  and  laying  out  lines  on  the  ground. 

Some  of  the  construction  principles  of  the  preceding 
experiments  may  be  modified,  or  used  without  modification 
in  practical  problems. 

EXERCISES 

1.  Show  how  the  angle  at  a  corner  of  the  room  may  be  bisected 
by  using  a  piece  of  "string  about  ten  feet  long. 

2.  Show  how  the  angle  of  a  field  may  be  bisected  by  using 

a  tape.  The  tape  must  be  so  used  that  it  is  unneces- 
sary to  draw  arcs  on  the  ground.  Points  may  be  marked 
by  stout  wire  marking  pins,  stuck  in  the  ground.  These 
should  be  made  of  about  &"  wire  and  be  about  12" 
long. 

3.  How  may  a  surveyor  bisect  the  length  of  a  curb  between  two 
street  corners? 

4.  How  may  a  surveyor  erect  a  perpendicular  to  a  line  by  using 
a  tape? 

5.  How  may  a  mason  erect  a  perpendicular  to  a  wall  by  using  a 
piece  of  string? 

6.  Adapt  the  construction  of  31  to  the  use  of  a  tape;  the  point 
being  nearer  the  line  than  the  length  of  the  tape. 

7.  How  may  a  surveyor  bisect  an  angle  by  means  of  a  transit? 

8.  How  may  a  surveyor  erect  a  perpendicular  by  means  of  a 
transit  ? 

9.  How  may  a  surveyor  lay  out  in  another  place  an  angle  equal 
to  a  given  angle  by  using  a  transit? 

10.  How  may  a  surveyor  lay  out  a  street  parallel  to  another  exist- 
ing street  by  making  use  of  the  principles  of  38  or  44?     Explain 


STRAIGHT    LINES,    CIRCLES,    SYMMETRY 


29 


at  what  point  an  angle  must  be  measured  with  a  transit,  and  at 
what  other  point  an  equal  angle  must  be  laid  out. 

11.  A  border  for  a  hardwood  floor  is  to 
be  laid  around  a  room  as  shown. 
How  may  the  angle  bisectors  be 
obtained  where  the  strips  of  wood 
are  fitted  together? 

12.  Sketch  a  symmetrical  design  for  a 
colored  glass  window  for  a  bath- 
room or  stairway. 

13.  Sketch  a  symmetrical  design  for  a 

decorated    plate,    book    cover,    lamp   shade,    rug,  wall-paper 
pattern,  or  some  other  familiar  object. 

59.  STANDARD  FORM  SHEET  FOR  RECORDING  FIELD  WORK. 
FIELD  WORK 


DEPARTMENT  OF  MATHEMATICS.- 


-SCHOOL 


•---.  c 


To  lay  out  a  line  parallel  to  the  curb  of  a  road  or  path,  and 
passing  through  a  given  point  (at  a  distance  of  several  hundred 
feet  from  the  curb) . 

Geometrical  Principle  Employed. — If  two  straight  lines 

are  cut  by  a  transversal  making  a  pair  of  alternate  interior 

£  angles  equal,  the  lines  are 

parallel. 

AB  is  the  curb  and  C  is 
the  given  point. 

(1)  Mark  any  convenient 
point  D,  on  the  curb,  and 
set  a  transit  over  point  D; 
measure  the  angle  BDC. 

(2)  Move  the  transit  to 
point  C;  lay  out  the  angle 

DCE  equal  to  the  angle  BDC',  drive  stake  E. 

(3)  CE  is  the  required  parallel. 

NOTE. — If  a  building  or  other  obstacle  is  in  the  way  of  the 
line  CE,  the  angle  DCF  may  be  laid  out  equal  to  180°- 
angle  BDC.  The  work  may  be  checked  by  repeating  with 
another  point  G  on  AB. 


CHAPTER  III 
PARALLELS,  PERPENDICULARS  AND  ANGLES 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

60.  The   Importance   of  Experimental  Investigation. — 
The  natural  world  about  us  is  full  of  facts.     We  learn  these 
facts  from  observation,  either  at  large  or  in  a  laboratory. 
Many  of  the  broader  principles  of  Chapter  I  may  be  con- 
sidered as  being  derived  from  general  observation.     We  see 
types  of  lines,  surfaces  and  solids,  symmetry  and  motion, 
indefinitely  repeated  in  natural  objects.     The  experimental 
results  which  were  obtained  in  Chapter  II  were  those  of  a 
geometrical  laboratory.     Every  science  passes  through  a 
certain  period  of  inception  during  which  facts  are  accumu- 
lated, and  this  is  true  of  chemistry,  physics,  psychology, 
biology,    physical   geography,    astronomy,    or  any  other 
science.     A   fact,    or   principle   is   proved  by  indefinitely 
repeating  the  experiment  or  observation. 

61.  The  Importance  of  Analysis  and  Classification.— 
Facts  alone  are  not  sufficient  to  constitute  a  science.     Ex- 
perimental   results    may    be    incorrectly    or    imperfectly 
stated,  due  to  errors  of  measurement  or  of  judgment.     If, 
however,  a  system  of  classification  can  be  built  up  in  which 
all  experimental  results  are  shown  to  agree  with  some  gen- 
eral principles,  and  thus  to  agree  among  themselves,  this 
very  agreement  is  a  further  proof  that  these  principles 
have    been    correctly    determined    by    experiment.     The 
greater  the  number  of  independent  principles  thus  success- 
fully analysed,  the  greater  will  be  the  certainty  that  the 
system  of  classification  adopted  is  a  true  expression  of 
nature's  laws.     We  will  now  proceed  with  the  analysis 
and  classification  of  the  facts  or  principles  which  we  have 
noted  in  Chapter  II. 

30 


PARALLELS,  PERPENDICULARS  AND  ANGLES      31 

62.  The  First  Step  in  Classification. — This  consists  in 
the  selection  of  the  simplest,  and  therefore  most  reliable, 
principles  as  indicated  by  the  experiments.     These  princi- 
ples are  either  incapable  of  analysis  in  terms  of  still  more 
elementary  principles,  or  are  not  readily  analyzed,  or  are 
so  elementary  that  it  is  not  worth  while  to  analyze  them. 

63.  Postulates. — These   are   the   elementary   principles 
selected  as  the  basis  of  classification.     Just  what  postulates 
are  sufficient  and  necessary  can  only  be  determined  as  the 
analysis  proceeds.     It  may  be  found  necessary  to  increase 
the  number  originally  selected  or  some  of  those  selected 
may  be  found  not  to  be  required. 

POSTULATES 

1.  Only  one  sect  can  be  drawn  between  two  given  points 
(24(6)). 

2.  Two  straight  lines  can  intersect  in  only  one  point 
(26(a)). 

3.  A  straight  line  and  a  circle,  or  two  circles,  can  inter- 
sect in  two  and  not  more  than  two  points  (26(6)  and  (c)). 

4.  A  straight  line  is  the  shortest  distance  between  two 
points. 

5.  A  sect  or  an  angle  can  be  divided  into  any  number  of 
equal  parts,  and  in  only  one  way  (28  and  29) . 

6.  All  straight  angles  are  equal. 

7.  All  right  angles  are  equal. 

8.  A  right  angle  is  one-half  of  a  straight  angle. 

9.  A  perpendicular  forms  a  right  angle. 

10.  Only  one  perpendicular  can  be  drawn  to  a  straight 
line  at  a  given  point  in  the  line,  or  from  a  given  point  with- 
out the  line  (30  and  31). 

11.  All  radii  of  the  same  circle  are  equal. 

12.  Any  definite  figure  can  be  constructed:  as  a  straight 
line  through  a  given  point;  a  line  through  two  given  points; 
a  sect  of  a  given  length;  an  angle  equal  to  a  given  angle;  a 
line  passing  through  a  given  point  and  perpendicular  to  a 
given  line;  etc.  (28  to  35,  38.) 


32  PLANE    GEOMETRY 

64.  Axioms. — These  are  numerical  relations  which  may 
also  enter  into  the  analysis  of  geometric  principles.     They 
are  general  in  their  application,  not  being  limited  to  geome- 
try, and  are  derived  from  reasoning  of  the  most  elementary 
kind. 

Complete  the  statement  of  the  axioms  which  are  left 
incomplete. 

AXIOMS 

1.  Things  equal  to  the  same  thing  are  equal  to  each  other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  subtracted  from  equals, —      — . 

4.  If  equals  are  multiplied  by  equals, —      — . 

5.  If  equals  are  divided  by  equals, —      — . 

6.  If  equals  are  added  to  unequals,  the  results  are  unequal 
in  the  same  order. 

7.  If  equals  are  subtracted  from  unequals, —      — . 

8.  If  unequals  are  subtracted  from  equals, —      — . 

9.  If  unequals  are  added  to  unequals,  the  greater  to  the 
greater  and  the  less  to  the  less,—      — . 

10.  If,  of  three  unequals,  the  first  is  greater  than  the 
second,  and  the  second  is  greater  than  the  third, —      — . 

11.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

12.  The  whole  is  greater  than  any  of  its  parts. 

13.  A  quantity  may  be  substituted  for  its  equal  in  any 
expression. 

65.  The  second  step  in  classification  is  the  preparation 
of  a  graded  list  of  those  experimental  results,  or  principles, 
which  can  be  analyzed  in  terms  of  other  principles,  and 
which  are  sufficiently  important  to  require  such  analysis 
and  explanation. 

66.  Theorems. — A  theorem  is  a  principle  that  is  analyzed 
in  more  elementary  terms. 

A  corollary  is  a  theorem  which  is  classified  .under  another 
theorem. 

The  order  of  arrangement  of  theorems  can  be  considerably 
modified.  All  that  is  necessary  is  to  preserve  their  con- 
secutive dependence. 


PARALLELS,    PERPENDICULARS    AND    ANGLES 


33 


anal. 

hypothesis 

hyp. 

ax. 

postulate 

post. 

con. 

theorem 

theo. 

cor. 

67.  Geometric    Style. — A    certain   formal   method    has 
been  developed  as  suitable  to  written  geometrical  analysis. 

The  statement  of  a  theorem  is  divided  into  two  parts : 

(1)  The  hypothesis,  which  is  a  statement  of  the  given,  or 
constructed,  properties  of  the  figure. 

(2)  The  conclusion,  which  is  a  statement  of  the  new,  or 
resulting,  properties  of  the  figure. 

The  analysis  (often  called  the  proof)  explains  how  the 
conclusion  is  worked  out  from  more  elementary  principles. 
It  consists  of  a  statement  of  each  separate  logical  step 
made,  and  the  reason  or  explanation  of  the  step. 

68.  Convenient  Abbreviations. 

analysis 
axiom 
conclusion 
corollary 

69.  Theorem  I. — Opposite  an- 
gles are  equal  (36). 

Hypothesis.  —  Straight  lines 
AB  and  CD  intersecting  and 
forming  pairs  of  opposite  angles, 
x  and  z,  y  and  w; 

Conclusion. — x  =  z,  y  =  w. 

Analysis. 

STATEMENT  REASON 

1.  x  +  y  =  a  straight  Z               Axiom  10 

2.  y  +  z  =  a  straight  Z               Axiom  11 

3.  x  +  y  =  y  +  z  Axiom    1 

4.  x  =  z  Axiom    3 

Show  also  that  the  fact  that  y  —  w,  is  dependent  upon 
these  axioms. 

EXERCISES 

1.  Draw  the  two  straight  angles  separately  which  are  considered 
as  making  up  the  composite  figure  of  the  theorem. 

2.  Does  the  analysis  show  that  the  fact  that  "opposite  angles  are 
equal"  is  not  an  independent  principle? 

3 


34  PLANE    GEOMETRY 

3.  Work  out  the  same  kind  of  analysis  for  x  =  z,  in    which  Zs 
x  and  z  are  combined  with    Z.w  instead  of  with    /.y. 

4.  In  analyzing  the  second  part,  y  =  w,  we  may  write  y  +  x  = 
a  straight    Z,  w  +  z  =  a  straight    /.,  y  +  x  =  w  +  z,  y=  w. 
Why  may  y  and  w  be  combined  with  different  angles  hi  the 
first  two  statements? 

5.  Upon  what  must  the  analysis  of  the  first  theorem  depend? 
Upon  what  may  that  of  the  second  theorem  depend?     Of  other 
theorems  ? 

ADDITIONAL  THEOREMS 

70.  These  are  theorems  of  minor  importance,  not  assigned 
a  theorem  number. 

(1)  Complements  of  equal  angles  are  equal. 

Helps. — Draw  two  equal  angles;  construct  their  comple- 
ments. State  the  hypothesis  and  conclusion.  Write 
the  analysis  to  show  how  the  equality  of  the  complements 
depends  upon  Postulate  7  and  Axioms  1,  3  and  11. 

(2)  Supplements  of  equal  angles  are  equal. 

71.  Theory  of  Parallels. — The  analysis  of  the  principles 
of  37,  and  39  to  46,  cannot  be  made  entirely  in  terms 
of  the  selected  postulates  and  previous  theorems.     Two 
additional  postulates  will  therefore  be  stated: 

Postulate  13. — Two  lines  that  are  cut  by  a  trans- 
versal making  a  pair  of  corresponding  angles  equal,  are 
parallel  (39). 

Postulate  14. — If  two  par- 
allel lines  are  cut  by  a  trans- 
versal, any  pair  of  correspond- 
ing angles  is  equal  (46). 

72.  Theorem  II. — Two  lines 
that  are  cut  by  a  transversal 
making    a   pair   of  alternate- 
interior  angles  equal,  are  paral- 
lel (40). 

Hypothesis. — AB  and  EF 
cut  by  transversal  CD  making 
alternate-interior  Zs  x  and  y 
equal : 


PARALLELS,  PERPENDICULARS  AND  ANGLES      35 

Conclusion.— AB\\EF. 
Analysis. 
STATEMENT  REASON 

1.  x  =  y  Hypothesis   (i.e.,   made  equal  in 

drawing  the  figure.) 

2.  x  =  z  Theorem  I  (a  principle  previously 

analyzed.) 

3.  y  =  z  Axiom  1. 

4.  AB\[CD  Postulate  13. 

EXERCISES 

1.  Draw  separate  figures  showing  the  opposite  angles  and  the  cor- 
responding angles  which  are  considered  as  making  up  the  figure 
of  the  theorem. 

2.  Make  the  analysis  using  the  corresponding  Zs  x  and  w. 

3.  Construct  a  figure  with  the  other  pair  of  alternate-ulterior  angles 
equal,  and  show  that  a  pair  of  corresponding  angles  is  also  equal, 
and  hence  the  lines  are  parallel. 

73.  Theorem  III. — Two  lines  that  are  cut  by  a  transversal 
making  a  pair  of  alternate-exterior  angles  equal,  are  parallel 
(41). 

Construct  the  figure;  state  the  hypothesis  and  conclu- 
sion. Analyze  (a)  by  using  Postulate  13 ;  and  (6)  by  using 
Theorem  II. 

ADDITIONAL  THEOREMS 

74.  (1)  State  the  result  of  the  experiment  of  42  as  a 
theorem.     Analyze  it  in  more  than  one  way. 

Helps. — (1)  x  -j-  y  =  a  straight  \  r 

Z;  why?—  (2)  x  +  z  =  a  straight  - 
Z;  Ax— (3)  z  +  y  =  x  +  z;  Ax  — 
(4)  y  =  z; — (5)  lines  are  ||. 

(2)  State  the  result  of  the  ex- 
periment  of  43  as  a  theorem.  \ 
Analyze  it  in  more  than  one  way. 

75.  Theorem  IV. — Two  lines  perpendicular  to  the  same 
line  are  parallel  (37). 

Helps. — (1)  The  right  angles  are  equal.     What  postulate? 


36  PLANE    GEOMETRY 

(2)  These  form  a  pair  of  angles  with  the  third  line  considered 
as  a  transversal. 

76.  A  converse  theorem  is  one  in  which  the  hypothesis 
and  conclusion  are  respectively  the  conclusion  and  hypo- 
thesis of  a  previous  theorem.     The   previous   theorem   is 
the  direct  theorem. 

77.  Theorem    V.     Converse    of    Theorem    II.— //   two 
parallel  lines  are  cut  by  a  transversal,  any  pair  of  alternate- 
interior  angles  is  equal  (46). 


State  the  hypothesis  and  conclusion. 
Analysis. 

STATEMENT  REASON 

1.  y  =  z  Postulate  14. 

2.  x  =  z  What  theorem? 

3.  x  =  y  What  axiom? 

EXERCISES 

1.  Draw  separate  figures  showing  how  the  figure  of  the  theorem  is 
made  up  of  more  elementary  parts. 

2.  Make  the  analysis  of  Theorem  V  with  some  other  pairs  of  alter- 
nate-interior and  corresponding  angles. 

78.  Theorem  VI.     Converse  of  Theorem  III  (46). 

State   the   theorem.     Draw   the   figure;   write   out   the 
analysis  in  more  than  one  way. 


PARALLELS,    PERPENDICULARS    AND    ANGLES 


37 


ADDITIONAL  THEOREMS 

79.  (1)  State  the  converse  of  Theorem  1,  74.     Analyze 
by  the  method  of  Theorem  V. 

(2)  State  the  converse  of  Theorem  2,  74.     Analyze  by 
any  method. 

80.  Theorem  VII.     Converse  of  Theorem  IV.— A  line 
which  is  perpendicular  to  one  of  two  parallels  is  perpendicular 
to  the  other  also  (46). 

Helps. — Consider   the    perpendicular   as   a   transversal. 

81.  Theorem  VIII. — Two  angles  whose  sides  are  parallel, 
each  to  each,  are  equal  if  both  are  acute  or  both  obtuse;  and  are 
supplementary  if  one  is  acute  and  one  obtuse  (48) . 


Hypothesis. — An  acute    /.x,  and  acute  Zs  y  and  z,  with 
sides  parallel  to  the  sides  of    Zx. 

Conclusion. — x  =  y  and  x  =  z. 

Helps. — (a)  Prolong  BA  and  EF  to  form  /.w.  Notice 
that  the  figure  now  contains  two 
pairs  of  parallel  lines  cut  by  trans- 
versals, and  hence  Zs  x  and  y  are 
both  equal  to  /.w.  Why?  .'.  x  =  y. 
Show  also  that  x  =  z. 

(b)  Draw    an   obtuse    angle,    and 
two  other  obtuse  angles  whose  sides 

are  parallel  to  the  sides  of  the  given  angle.     Show  that 
these  angles  are  each  equal  to  the  given  angle. 

(c)  Draw  an  acute  angle,  and  two  obtuse  angles  whose 


38 


PLANE    GEOMETRY 


sides  are  parallel  to  the  sides  of  the  given  angle.     Show 
that  these  angles  are  supplementary  to  the  given  angle. 

82.  Theorem  IX. — Two  angles  whose  sides  are  perpendicu- 
lar, each  to  each,  are  equal  if  both  are  acute  or  both  obtuse, 
and  are  supplementary  if  one  is  acute  and  the  other  obtuse 
(49). 


State  the  hypothesis  and  conclusion. 
Analysis. 

REASON 
What  postulate? 


What  theorem? 
What  theorem? 
70.  Theorem  (1) 
What  axiom? 


STATEMENT 

1.  Draw  BG1.BA  and  BE 
±BC 

2.  BG\\ED  and  BH\\EF 

3.  w  =  y 

4.  w  =  x 

5.  .'.  x  =  y 

Analyze  also  for  other  angles  as  in  Theorem 
VIII  (81). 

83.  General  directions  to  be  observed  in  the  analysis  of 
theorems. 

1.  Draw  large  and  distinct  figures,  with  all  necessary 
and  no  unnecessary  lettering. 

2.  Construct  the  figure  from  its  given  properties. 


PARALLELS,  PERPENDICULARS  AND  ANGLES 


39 


3.  Eliminate  all  useless  statements. 

4.  Cultivate  a  correct  style. 

5.  Auxiliary  lines  may   be   drawn  with  any   assigned 
property.     If  the  line  possesses  additional  properties,  it  is 
part  of  the  analysis  to  show  that  these  exist.     Such  lines 
are  drawn  dotted. 


ADDITIONAL  THEOREMS 

84.  (1)  The  bisectors  of  two  supplementary-adjacent  angles 
are  perpendicular. 

(2)  An  angle  ABC  is  bisected  by  BD;  through  the  vertex 
B  a  line  EF  is  drawn  perpendicular  to  BD.    Line  EF  makes 
equal  angles  with  the  sides  of  angle  ABC. 

State  this  theorem  in  general  terms. 

(3)  In  AABC,  AB>AC;  point  D  is  taken  on  AB  so  that 
AD  =  AC.     ThenBOBD. 

(4)  The  bisectors  of  alternate-interior  angles  of  parallel 
lines  are  parallel. 

(5)  The  bisectors  of  corresponding  angles  of  a  system  of 
parallels  form  another  system  of  parallels. 

(6)  The  bisectors  of  two  pairs  of  opposite  angles  formed  by 
two  intersecting  straight  lines  are  perpendicular. 

(7)  The  bisectors  of  a  pair  of  consecutive-interior  angles  of 
parallels  are  perpendicular. 

Helps.— (1)  Draw  MN\\EF.—(2)  x  +  y  =  a  straight  Z; 
—(3)  s  +  t  =  a,  right  /;—  (4)  s  =  v,  t  =  w,  etc. 


40 


PLANE    GEOMETRY 


(8)  Two  lines  parallel  to  the  same 
line  are  parallel  to  each  other  (47). 

Helps.— Hyp.:  AB  and  BF\\CD. 
Analysis. — (1)  DrawG#;  (2)  x  = 
y,y=z;  (3)  x  =  z\  (4)  AB\\EF. 

(9)  //    one    straight    line    meets 
another  straight  line  so  that  adjacent 
angles    are    equal,    the    angles    are 
right  angles,  and  the  lines  are  per- 
pendicular. 

Helps.— (1)  x  +  y  =  180°;— 
(2)  x  =  y;—(3)2x  =  180°;  — (4>  x 
=  90°;— (5)  CD.LAB. 


85.  REVIEW  EXERCISES 

1.  Are  we  compelled  to  follow  any  certain  order  in  making  experi- 
ments?    Or  are  the  experiments,  for  the  most  part,  completely 
independent  of  each  other?     Are  the  analyses  of  the  results  of 
these  experiments  also  independent? 

2.  Mark  two  points  on  the  surface  of  a  globe,  or  ball.     Will  the 
straight  line  containing  them  lie  wholly  in  the  surface? 

3.  Draw  a  line  on  the  paper  by  marking  along  an  edge  of  your 
triangle.     Turn  the  triangle  over  on  this  line  as  an  axis.     Does 
the  drawn  line  still  coincide  with  the  edge  of  the  triangle?     If 
so,  what  kind  of  line  is  it?     Has  any  other  line  this  property? 

4.  How  many  straight  lines  are  determined  by  three  points? 

5.  How  many  points  are  determined  by  three  straight  lines? 

6.  Do  two  points  in  space  determine  a  straight  line? 

7.  Of  what  angle  is  the  complement  equal  to  the  angle? 

8.  Of  what  angle  is  the  supplement  equal  to  the  angle? 

9.  Of  what  angle  is  the  supplement  double  the  angle? 

10.  The   difference   of   two   supplementary   angles   is   d   degrees. 
Find  the  angles. 

11.  Given  a  line  AB,  a  point  C  without  it,  and  an  angle  x.     Find 
a  point  D  on  AB  such  that  CD  makes  an  angle  CD  A  =  x. 
Use  the  rule  and  compass  method. 

12.  x  and  2  x  represent  any  pair  of  adjacent  angles   on  a  figure 
formed  by  a  transversal  cutting  two  parallels.     Draw  the  figure 
and  write  on  it  the  measures  in  degrees  of  the  eight  angles. 

13.  Two   parallels  are  cut  by  a  transversal  so  that  x  and  y  are 


PARALLELS,  PERPENDICULARS  AND  ANGLES 


41 


adjacent  angles  and  x  and  2y  are  corresponding  angles.     Sketch 
the  figure  and  calculate  all  the  angles  in  degrees. 

14.  Draw  the  right  angle  of  your  triangle.     Turn  the  triangle  over 
using  a  perpendicular  side  as  an  axis.     Draw  the  right  angle 
again  adjacent  to  the  angle  first  drawn.     How  does  this  draw- 
ing test  the  accuracy  of  the  right  angle  of  the  triangle? 

15.  The  sum  of  two  adjacent  angles  is  85°.     Find  the  angle  between 
their  bisectors? 

16.  The  angle  between  the  bisectors  of  two  adjacent  angles  is  80° 
30'.     What  is  the  sum  of  the  angles? 

17.  Reproduce    this    figure    on  a  larger    scale. 
In  what  does  the  illusion  consist? 

18.  Measure  the  distance  from  the  vertex  of  the 
right  angle  of  your  triangle  to  the  hypotenuse. 

19.  A  country  road  runs  from  southeast  to  northwest.     In  what 
direction  would  you  cross  the  fields  from  a  point  to  the  west  of 
the  road,  to  reach  it  by  the  shortest  way? 

20.  Would  surveyors,  architects  and  navigators  be  able  to  make 
use  of  the  principles  of  geometry  even  if  the  analysis  of  experi- 
mental principles  had  never  been  developed? 

21.  In  what  two  ways  are  axioms  distinguished  from  postulates? 

22.  John  and  James  each  received  two  dollars  for  spending  money 
during  a  holiday.     John  spent  more  of  his  money  than  James 
spent  of   his.     Who   has   more   remaining?     What  axiom   is 
illustrated? 

23.  If  the  greater  quantity  of  one  pair  of  unequals  is  added  to  the 
less  quantity  of  another  pair  of  unequals,  and  the  less  quantity 
of  the  first  pair  is  added  to  the  greater  of  the  second  pair, 
which  sum  will  be  greater?     Can  such  a  result  be  stated  as  an 
axiom  ? 

24.  Express   the   axioms   algebraically.     Thus,    Axiom    1;   a  =  b, 
a  =  c,  . ' .  b  =  c. 

25.  Which  of  the  axioms  have  been  used  in  algebra:  (1)  In  the  re- 
duction of  equations,   (2)  in  the  processes  of  elimination  in 
simultaneous  equations,  (3)  in  the  reduction  of  inequalities? 

APPLICATIONS 

86.  LIGHT  REFLECTED  FROM  MIRRORS. 
The  law  of  reflected  light:  It  is  proved 
experimentally  that  the  angle  of  inci- 
dence, x,  equals  the  angle  of  reflection, 
y,  or  x'  =  yf.  NN  is  called  the  normal 
to  the  plane  of  the  mirror  MM. 


42 


PLANE    GEOMETRY 


EXERCISE 


Show  that  if  two  mirrors, 
M  and  Mr,  are  parallel,  the 
incident  ray  A  and  the  finally 
reflected  ray  B,  are  parallel. 
Help. — Show  that  a  =  /3. 


87.  SURVEYING. — Measurement  of  an  inaccessible  angle. 
An  angle  in  an  inaccessible  corner  of  a  field  may  be  measured 
as  shown  in  the  sketches.  Ex- 
plain how  this  is  done  and  the 
principle  involved. 


Bearings. — Point  0  is  a  point    of   ob- 
servation; SN  is  a  meridian  through  0; 

the    angles    x,  y,  z,  w,  are  the  bearings  B 

of  points  Ay  B,  C,  D,  as  seen  from  0.    A 

is  northeast  of  0,   B  is    northwest,   etc. 

If  x  =  35°,  the  bearing  of  line  OA  is  read 

north  35  degrees  east,  and  is  written  N 

35°  E.     The  length  OA  is  the  distance  or 

course.     A  point   can   be  located  from  a 

given  point  when  its  bearing  and  distance  and   also   the 

direction  of  the  merid- 
ian through  the  given 
point  are  known. 

A  surveyor  may  make 
a  map  of  a  piece  of 
ground,  or  of  a  broken 
line,  by  measuring  bear- 
ings and  distances,  first 
at  point  A,  next  at  point 
B,  etc.  . 


PARALLELS,  PERPENDICULARS  AND  ANGLES 


43 


An  angle  of  elevation  is  an  angle  measured  in  a  vertical 
plane  between  a  horizontal  and  an 
oblique    line,    the  horizontal   line 
being  below  the  oblique  line. 

An    angle    of   depression   is   an 
angle    measured   in   a   vertical 
plane  between  a  horizontal  and 
an  oblique  line,  the  horizontal 
line  being  above  the  oblique  line 

EXERCISES 

1.  An  imaginary  line  is  drawn  from  the  top  B  of  a  hill  to  a  point  A 
in   the  plain   that  stretches 

away  from  its  base.  The 
angle  of  elevation  x  of  the 
top  of  the  hill  as  observed  at 
A  is  35°.  What  is  the  angle 
of  depression  y  of  the  point 
A  as  observed  at  £?  Why? 

2.  Are   two   vertical  lines,   determined   by   plumb-lines 
several  feet  or  several  hundred  feet  apart  absolutely 
parallel?     Why  not? 

3.  Draw   any  meridian   SN  through  a  given  point  X. 
Locate  a  point  Y  whose  bearing  from  X  is  S  50°  W, 
and  whose  distance  from  X  is  175  feet.     Plot  on  a 
scale  of  100  feet  to  the  inch. 

4.  The  bank  of  a  river  is  located  by  a  series  of  straight  lines.     Plot 
the  river  bank  by  drawing  an  irregular  curved  line  agreeing 
closely  with  the  broken  line  ABCDEF.    Such  a  line  is  called  in 
surveying  a  traverse  line. 

FIELD-NOTES 


Beginning  point 
of  line 

Bearing  of  line  from 
beginning  point 

Length  of  line, 
feet 

A        ... 

N  50°  E 

120 

B  

N72°E 

50 

c 

S  45°  E 

150 

D  

S  80°  W 

70 

E.. 

N25°  W 

120 

44 


PLANE    GEOMETRY 


88.  LATITUDE. — The  latitude  of  a  place  is  determined  by 
measuring  the  angle  of  elevation  e,  of  the  sun  when  it  is 
directly  south,  which  occurs  at,  or  about,  noon. 

The  angle  of  elevation  of  the  sun  can  be  most  accurately 
determined  by  using  an  artificial  horizon.  This  is  a  mirror 
accurately  levelled,  or  a  shallow  disk  of  mercury;  Af, 
Figure  2. 

S 


In  Figure  3,  P  is  the  place  on  the  earth  of  which  the 
latitude  is  required,  EQ  is  the  equator,  NN'  is  the  axis  of 
the  earth,  PH  is  the  horizontal  plane  corresponding  to  TH 
of  Figures  1  and  2.  The  angle  QOP  is  the  latitude  of  P. 
This  figure  shows  the  position  of  the  sun  with  respect  to 
the  earth  at  the  instant  of  equinox,  which  occurs  at  some 
place  on  the  earth  on  March  21st  and  September  23d. 

At  all  times  between  September  23d  and  March  21st, 


PARALLELS,  PERPENDICULARS  AND  ANGLES 


45 


FIG.  5. 


the  sun  is  south  of  the  equator,  and  the  angle  which  a 

line     OS',    or    the     (approximately) 

parallel  line  PS,  makes  with  the  plane 

EQ  of  the  earth's  equator,  is  called 

the  south  or  minus  declination  of  the 

sun.     This  angle  is  QOS'. 

At  all  times  between  March  21st 
and  September  23d,  the  sun  is  north 
of  the  equator,  and  the  angle  QOS'  A 
is  called  its  north  or  plus  declination, 
Figure  5. 

Commander  Peary,  Capt.  Amund- 
sen and  Capt.  Scott  located  the  poles 
of  the  earth  by  this  or  a  similar 
method. 

EXERCISES 

1.  Show  that  one-half  of  the  total  angle  STM,  Figure  2,  is  the 
angle  of  elevation  of  the  sun,  STH. 

2.  How  is  the  angle  of  elevation  E,  Figure  3,  related  to  the  latitude 
angle  L,  of  place  P? 

3.  In  Figure  4,  draw  OA  _J_  OS'.     Show  that  the  declination  angle 

D,  of  the  sun,  equals  angle  NO  A;  and  that  the  angle  of  elevation 

E,  of  the  sun,  equals  angle  AOP.     Then  L  =  90°  -  (E  +  D); 
and  if  ZD  is  given  its  minus  sign,  L  =  (90°  —  E)  +  D. 

4.  In  Figure  5,  draw  OA  JL  OS'.     Show  that  the  declination  angle 
D   =    ANOA;  and  that    £E   =    /.AOP.     Then  L   =  90°  + 
D  -  E;  and  if  D  is  given  its  plus  sign,  L  =  (90°  -  E}  +  D,  as 
in  Exercise  3. 

5.  If  the  angle  of  elevation  of  the  sun  measured  near  the  Old  State 
House  in  Philadelphia,  at  an  instant  when  the  sun  was  exactly 
south,  and  when  its  declination  was  zero,  was  50.05°;  what  is 
the  latitude  of  this  point? 

6.  The  angle  of  elevation  of  the  sun  when  its  direction  is  south, 
measured  at  a  place  P,  was  75°  10'.     The  declination  at  this 
time  was    +13°  45'.     What   is   the  latitude  of  the  place  of 
observation? 

Another  Method  of  Finding  the  Latitude  of  a  Place. — The 
measuring  quadrant  illustrated  in  Figure  6  was  first  used 
by  Tycho  Brahe,  a  Danish  astronomer  (1546-1601),  to 
measure  the  angles  of  elevation  of  stars.  If  the  angle  of 


46 


PLANE    GEOMETRY 


elevation  HPS  of  the  pole  star  (Polaris)  is  measured  at  any 
time  of  the  year,  (Figures  6  and  7)  the  latitude  L  of  the 
place  P  can  be  found.  The  reason  of  this  is  that  the  pole 
of  the  earth  extended  northwards  to  its  point  of  intersection 
with  the  "  celestial  sphere  "  is  very  close  to  Polaris.  Polaris, 
however,  describes  a  small  circle  around  the  pole,  and  there- 
fore the  angle  of  elevation  of  the  star  must,  in  general,  be 
corrected  to  obtain  the  angle  of  elevation  of  the  pole.  This 

s. 


FIG.  7. 

method  is  troublesome  to  apply  because  the  crosshairs  of 
the  telescope  must  be  illuminated,  since  the  observation  is 
made  at  night.  The  correction  of  the  angle  requires  the 
use  of  certain  tables  published  by  the  Government  Printing 
Office  at  Washington,  D.  C. 

EXERCISES 

7.  Show  what  arc  of  the  quadrant,  Figure  6,  is  read  to  give  the 
value  of  angle  HPS,  and  show  what  relation   /.HPS  has  to  the 
latitude  angle  L,  of  Figure  7. 

8.  The  angle  of  elevation  of  Polaris  is  measured  at  a  certain  place 
=  49°  18'.     The  correction  to  be  applied  to  obtain  the  angle  of 
elevation  of  the  pole  is  —  1°  2'.     Find  the  latitude  of  the  place 
of  observation. 

FIELD  WORK 

89.  DEPARTMENT  OF  MATHEMATICS —     — SCHOOL. 

To  determine  the  latitude  of  School. 

Geometrical  principle  employed: 

Two  acute  angles  whose  sides  are  perpendicular,  each  to 
each,  are  equal. 


PARALLELS,  PERPENDICULARS  AND  ANGLES      47 

METHOD. — (1)  Set  up  and  accurately  level  the  transit, 
about  15  minutes  before  noon  on  March  21st,  or  September 
23d. 

(2)  Set  the  horizontal  crosshair  tangent  to  the  lower 
edge  of  the  sun;  as  the  angle  of  elevation  of 

the  sun  increases,  follow  it  up  with  the 
crosshair,  until  the  angle  of  elevation  has 
reached  its  maximum. 

(3)  Read  the  vertical  angle;  level  the  telescope  by  the 
telescope  spirit  level  and  read  the  index  error  of  the  vertical 
circle. 

(4)  Add  0°  15'  to  the  value  of  the  angle  of  elevation,  to 
allow  for  parallax,  refraction  and  the  radius 

of  the  sun. 

(5)  Latitude  =  90°  —  corrected  angle  of 
elevation  of  sun. 

Note:  Dark  glasses  must  be  fastened  in  front  of  the 
object  glass  of  the  telescope. 

To  determine  the  latitude  of  a  place  with  accuracy  within 
one  or  two  minutes,  the  Nautical  Almanac  must  be  used. 

The  latitude  of  a  place  can  be  determined  on  any  day  of 
the  year,  if  the  meridian  altitude  of  the  sun  is  measured  on 
that  day.  and  the  tables  found  in  the  Nautical  Almanac  for 
decimation,  parallax,  refraction  and  semi-diameter  of  the 
sun,  are  consulted. 


CHAPTER  IV 
TRIANGLES 

PRINCIPLES  DETERMINED  EXPERIMENTALLY 

90.  Triangles   Defined. — A   triangle   is   a   plane   figure 
bounded  by  three  straight  lines. 

The  sides  are  the  bounding  lines. 

The  perimeter  is  the  sum  of  the  sides. 

The  interior  angles,  or  the  angles,  of  the  triangle  are  the 
angles  formed  by  the  sides. 

The  vertices  of  the  triangle  are  the  vertices  of  the  angles. 

An  exterior  angle  is  an  angle  formed  by  one  of  the  sides 
extended  and  the  adjacent  side  not  extended. 

The  parts  of  a  triangle  are  the  three  sides  and  the  three 
angles. 

B 


EXERCISES 

1.  Name  the  triangle  of  the  figure.     Name  the  sides;  the  angles  by 
a  single  letter;  the  vertices;  the  exterior  angles  which  are  drawn. 

2.  What  angle  of  the  figure  is  neither  an  interior  nor  an  exterior 
angle?     To  which  angle  is  it  equal? 

3.  Draw  a  triangle  with  all  possible  exterior  angles. 

4.  What  is  the  relation  between  an  interior  and  an  exterior  angle 
at  the  same  vertex? 

91.  Triangles    Classified. — The    classification    is    based 
upon:  (1)  the  relative  length  of  sides;  (2)  the  size  of  angles. 

48 


TRIANGLES  49 

According  to  sides,  a  triangle  is  called: 

Equilateral,  when  the  three  sides  are  equal. 

Isosceles,  when  two  sides  are  equal. 

Scalene,  when  no  two  sides  are  equal. 
According  to  angles,  a  triangle  is  called : 

Equiangular,  when  the  three  angles  are  equal. 

Right,  when   one  angle  is  a  right  angle. 

Obtuse,  when  one  angle  is  obtuse. 

Acute,  when  all  angles  are  acute. 

Oblique,  when  it  is  not  a  right  triangle. 

EXERCISES 

1.  Draw  triangles  of  each  of  the  eight  kinds. 

2.  Draw  triangles  of  every  possible  kind,  combining  a  name  from 
the  first  list  with  one  from  the  second;  as,  obtuse-isosceles;  etc. 

3.  What  triangle  is  the  general  type! 

92.  Relations  Between   Triangles. — Congruent  triangles 
are  triangles  which  are  alike  in  all  respects,  and  which  can 
be  made  to  coincide. 

Similar  triangles  have  the  same  shape,  or  angles,  but  not 
the  same  size,  or  lengths  of  sides.  They  are  considered  in 
Chapters  X  and  XI. 

The  homologous  parts  of  congruent  or  similar  triangles  are 
the  correspondingly  situated  parts. 

EXERCISES 

1.  Draw  two  congruent  triangles.     Name  all  the  homologous  sides, 
angles,  vertices,  and  exterior  angles. 

2.  Draw  two  similar  right  triangles.     Name  the  homologous  sides 
and  angles. 

93.  Position  of  the  Parts  of  a  Triangle. 

An  angle  is  included  by  two  sides. 
A  side  is   included  by  two  angles. 
Two  angles  are  adjacent  to  each  side. 
Each  side  is  opposite  an  angle. 
Each  angle  is  opposite  a  side. 


50  PLANE  GEOMETRY 

EXERCISES 

1.  Draw  a  triangle  ABC;  letter  the  sides  opposite  the  angles  A,  B,C, 
respectively  a,  6,  c.     This  will  be  called  the  standard  lettering  of 
a  triangle. 

2.  What  angle  is  included  by  a  and  c?     What  sides  include  C? 
What  angles  are  adjacent  to  a;  to  6?     What  angle  is  opposite  6? 
What  side  is  opposite  C?     What  side  is  included  by  angles  A 
andC? 

Names  of  Special  Parts  of  Triangles. 
In  an  isosceles  triangle: 

The  base  is  the  unequal  side. 

The  vertex  angle  is  included  by  the  equal  sides. 

The  base  angles  are  adjacent  to  the  base. 
In  a  right  triangle : 

The  perpendicular  sides  include  the  right  angle. 

The  hypotenuse  is  opposite  the  right  angle. 

EXERCISES 

1.  Draw  an  isosceles  triangle.     Name  all  the  parts. 

2.  What  angle  of  the  triangle  in  Exercise  1  is  opposite  the  base? 
What  angles  are  opposite  the  equal  sides? 

3.  Draw  a  right  triangle.     Name  all  the  parts. 

4.  What  angles  of   the  right  triangle   include   the  hypotenuse? 
What  angles  are  opposite  the  perpendicular  sides? 

In  any  triangle: 

The  base  is  the  side  on  which  it  stands;  but  any  side 

may  be  called  the  base. 
•     An  altitude  is  a  line  drawn  from  a  vertex  perpendicular 

to  and  meeting  the  opposite  side. 
A  median  is  a  line  drawn  from  a  vertex  to  the  middle 
point  of  (bisecting)  the  opposite  side.' 

EXERCISES 

5.  Draw  a  triangle  with  all  possible  altitudes. 

6.  In  what  kind  of  triangle  does  one  altitude  (or  more)  fall  upon  a 
side  produced? 

7.  In  what  kind  of  triangle  are  some  of  the  sides  the  altitudes? 

8.  Draw  a  triangle  with  all  possible  medians.     Will  the  medians  of 
a  triangle  ever  fall  outside  the  triangle? 


TRIANGLES  51 

Abbreviations. 

altitude alt.  isosceles isos. 

congruent,  is  congruent  to . .  ^  similar ~ 

homologous horn.        triangle,  triangles A,  A 

94.  Experiment  I. — Properties  of  the  angles  of  a  triangle. 

(a)  The  number  of  obtuse  angles  possible  in  a  triangle. 
(6)  The  number  of  right  angles  possible  in  a  triangle. 

(c)  The  number  of  acute  angles  possible  in  a  triangle. 

(d)  The  sum  of  the  angles  of  a  triangle. 

(e)  The   relation   between   an   exterior   angle   and   the 
interior  angles  of  a  triangle. 

Part  (a).-^-(l)  Draw  a  triangle  with  one  obtuse  angle. 
(2)  Draw,  if  possible,  a  triangle  with  more  than  one  obtuse 
angle.  State  result. 

Part  (6). — Make  the  same  experiment  for  one  or  more 
right  angles.  Notice  that  an  infinite  triangle  can  be  drawn 
with  two  right  angles.  State  result. 

Part  (c). — How  many  acute  angles  may  a  triangle  have? 
How  many  must  it  have?  State  result. 

Part  (d). — Measure  the  angles  of  several  triangles  of  diff- 
erent shapes  with  the  protractor.  State  result. 

Part  (e). — Measure  the  exterior  angles  of  the  triangle  of 
part  (d)  and  compare  their  values  with  the  sum  of  certain 
interior  angles.  State  result. 

95.  EXPERIMENT  II. — A  method  of  finding  the  third  angle 
of  a  triangle  when  two  angles  are  known,  without  drawing 
the  triangle. 


METHOD  (a). — Draw  the  two  given  angles,  x  and  y. 
Draw  the  angles  x  and  y  adjacent,  and  find  the  angle  z 
which  completes  a  straight  angle,  z  is  the  third  angle 
required. 

METHOD  (6). — Measure  the  given  angles  with  a  pro- 
tractor, and  find  the  third  angle  without  drawing  it. 


52  PLANE    GEOMETRY 

EXERCISES 

1.  Two  of  the  angles  of  a  triangle  are  75°  and  38°.     Calculate  the 
third  angle  without  a  drawing. 

2.  Can  a  triangle  be  drawn  with  the  angles  81°,  47°,  52°?     With 
the  angles  35°,  120°,  40°?     With  the  angles  20°,  136°,  12°? 

3.  Can  a  triangle  be  drawn  with  two  (interior)  angles  75°  and  45° 
and  the  exterior  angle  at  the  third  vertex  equal  to  120°. 

4.  If  the  interior  angles  A,  B,  C  of  a  triangle  are  70°,  30°  and  80°, 
calculate  the  three  exterior  angles  A',  B',  Cf,  adjacent  respectively 
to  angles  A,  B,  C. 

5.  If  all  the  angles  of  a  triangle  are  equal,  how  large  is  each? 

6.  If  one  angle  of  a  triangle  is  126°,  and  the  other  two  angles  are 
equal,  how  large  is  each  exterior  angle? 

7.  If  two  angles  of  a  triangle  are  each  72°  30',  calculate  the  third 
angle. 

8.  One  angle  of  a  triangle  is  double  another  angle,  and  the  third 
angle  is  20°  more  than  the  smallest  angle.     Calculate  the  angles 
and  sketch  the  triangle. 

96.  Construction  of  Triangles  from  Given  Parts. — In 
general,  three  given  parts,  sides  or  angles,  are  sufficient  to 
construct  a  triangle.  If  less  than  three  parts  are  given, 
the  triangle  is  not  determined.  If  more  than  three  parts 
are  given,  the  triangle  cannot  be  constructed  unless  the 
additional  parts  agree  by  chance  or  by  calculation  with  the 
triangle  as  determined  by  three  given  parts. 

Different  Combinations  of  Three  Parts  of  a  Triangle. 

(1)  Three  sides. 

(2)  Two  sides  and  one  angle;  of  which  there  are  three 
possible  arrangements: 

(a)  Two  sides  and  the  included  angle. 

(&)  Two  sides  and  the  angle  opposite  the  greater  of  the 

two  sides, 
(c)  Two  sides  and  the  angle  opposite  the  less  of  the 

two  sides. 

(3)  One  side  and  two  angles;  of  which  there  are  three 
possible  arrangements: 

(a)  The  side  included  by  the  two  angles. 
(6)  The  greater  angle  opposite  the  side  and  the  less 
angle  adjacent  to  it. 


TRIANGLES  53 

(c)  The  less  angle  opposite  the  side  and  the  greater 
angle  adjacent  to  it. 

(4)  Three  angles. 

Values  of  the  Given  Parts. — The  lengths  of  sides  may  be 
given  (1)  equal  to  sects  which  are  drawn,  or  (2)  by  numerical 
measure.  The  size  of  angles  may  be  given  (1)  equal  to 
drawn  angles,  or  (2)  by  numerical  measure. 

97.  Experiment  III. — A  method  of  drawing  a  triangle  of 
which  three  sides  are  given. 

Given. — Sects  a,  6,  e,  from 
which  the  sides  of  the  triangle 
are  to  be  measured.  b 


METHOD. — (1)  Measure  CB  =  a. 

(2)  Draw  two  arcs,  centers  at  C  and  B,  with  radii  equal 
respectively  to  b  and  c,  intersecting  each  other  at  A. 

(3)  Draw  AC  and  AB. 

(4)  ABC  is  the  required  triangle. 

Discussion. — (1)  If  a  =  b  +  c,  the  triangle  becomes  a 
straight  line. 

(2)  If  a<b  +  c,  no  triangle  is  possible. 

EXERCISES 

1.  Does  the  order  of  placing  the  sides  affect  the  shape  or  size  of  the 
triangle? 

2.  Construct  a  triangle  whose  sides  are  3.2,  2.4,  and  2.7  inches. 
Measure  the  first  side  from  the  scale,  and  measure  the  two  radii 
from  the  scale.     Measure  the  angles  of  the  triangle. 

3.  Are  triangles  possible  whose  sides  are  (1)  12,  17,  15;  (2)  106,  363, 
257;  (3)  93,  71,  50;  (4)  28,  28,  15;  (5)  80,  80,  80;  (6)  29,  81,  46? 

4.  Construct  an  isosceles  triangle  whose  sides  are  3,  3  and  4  inches. 

5.  Construct  an  equilateral  triangle  whose  sides  are  each  3.5  inches. 

98.  Experiment  IV.— To  draw  a  triangle  of  which  two 
sides  and  the  included  angle  are  given. 

Helps. — Draw  two  sects  a  and  b,  and  an  angle  C, 
from  which  the  given  parts  of  the  triangle  are  measured. 


54  PLANE   GEOMETRY 

Describe  the  method  as  in  97.  Are  there  any  special  or 
impossible  cases? 

EXERCISES 

1.  Construct  a  triangle  of  which  two  sides  are  2.9  and  1.5  inches, 
and  the  included  angle  is  57°.     Measure  the  sides  from  the  scale 
and  the  angle  directly  from  the  protractor.     Measure  the  remain- 
ing side  and  angles. 

2.  Are  triangles  possible  with  the  parts  (1)  a  =  150,  b  =3,  C  = 
86°;  (2)  a  =  10,  b  =  10,  C  =  162°;  (3)  a  =  7,  c  =  5,  B  =  215°? 

3.  Construct  an  isosceles  triangle  whose  equal  sides  are  each  3.0 
inches  and  whose  vertex  angle  is  70°. 

4.  Construct  a  right  triangle  whose  perpendicular  sides  are  4.2  and 
1.7  inches. 

99.  Experiment  V. — To  draw  a  triangle  of  which  two 
sides  and  the  angle  opposite  the  greater  of  the  two  sides 
are  given. 

Construct  the  figure,  and  describe  as  in  97.     Note  any 
special  or  impossible  cases. 

EXERCISES 

1.  Construct  a  triangle  of  which  two  sides  are  2.9  and  1.5  inches, 
and  the  angle  opposite  the  2.9-inch  side  is  57°.     Measure  the 
remaining  parts. 

2.  Construct  a  right  triangle  of  which  the  hypotenuse  is  4.0"  and 
a  perpendicular  side  is  2. 3". 

100.  Experiment  VI. — To  draw  a  triangle  of  which  two 
sides  and  the  angle  opposite  the  less  of  the  two  sides  are 
given.  a -i — +- 

This  is  the  most  com-     A"1 
plicated  construction  of 
a     triangle    from   three 
given     parts.       Several 
cases  occur. 

(1)   Use    these    given 
measurements.        Con-  FlG-  1- 

struct  two  entirely  different  triangles  (two  figures)  from 
the  measurements,  the  required  relative  position  being 
maintained  in  both  figures. 


TRIANGLES 


55 


(2)  Use    these    given 
measurements .      What 
do  you  observe   in   this 
case  as  to  the  number  of 
possible    triangles    and 
the  kind  of  triangles  ob- 
tained? 

(3)  Use    these    given 
measurements.      What 

do  you  observe  in  this  case? 


B 


FIG. 


FIG.  2. 

(4)  Draw  any  two  sects 
a  and  6,  with  a>6,  and  a 
right  angle   B.    Construct 
the  triangle,  if  possible. 

(5)  Draw  any  two  sects 
a  and  6,  with  a>6,  and  an 
obtuse  angle  B.    Construct 
the  triangle,  if  possible. 

In  order  to  classify  these 
results,  draw  a  perpendicular  in  Figures  (1),  (2)  and  (3), 
from  vertex  C  of  the  triangle,  to  side  c,  as  in  Figure  4. 
Call  this  the  perpendicular  distance,  p,  of  the  triangle. 

RESULT. — When  the  given  angle  is  acute  and  (1)  when 
b>p,  two  different  triangles  can  be  constructed;  (2)  when 
b  =  pt  one  right  triangle  can  be  constructed;  (3)  when 
b<p,  no  triangle  can  be 
constructed. 

When  the  given  angle  is 
right  or  obtuse  (the  angle 
being  opposite  the  less  of 
the  two  given  sides),  no 
triangle  can  be  constructed. 

Because  of  the  fact  that. 
two  entirely  different  tri-  B 
angles    can  sometimes  be 
constructed  from  the  same  measurements,  this  construction 
is  called  the  ambiguous  case  in  triangles. 


\p 


FIG.  4. 


56  PLANE  GEOMETRY 

EXERCISES 

Construct  triangles  from  the  following  measurements.  Note  which 
case  arises  in  each  construction.  Measure  the  remaining  parts  in 
each  figure  (in  some  figures  there  are  two  sets  of  parts). 

1.  a  =  3.2",  b  =  2.1",  B  =  25°     4.  a  =  3.2",  b  =  1.5",  B  =    40° 

2.  a  =  3.2",  b  =  2.1",  B  =  50°     5.  a  =  2.7",  b  =  2.0",  B  =    90° 

3.  a  =  3.2",  b  =  1.6",  B  =  30°     6.  a  =  2.9",  6  =  1.8",  B  =  110° 

101.  Experiment  VII. — To  draw  a  triangle  of  which  two 
angles  and  the  included  side  are  given. 

Construct  the  figure  from  given  parts  and  observe  special 
and  impossible  cases. 

EXERCISES 

1.  Construct  a  triangle  of  which  one  side  =  2.8",  and  the  two  angles 
including  the  side  are  42°  and  70°. 

2.  Construct  an  isosceles  triangle  of  which  the  base  is  4.0"  and  the 
base  angles  are  each  35°. 

3.  Which  of  the  following  triangles  are  (a)  possible,  (6)  special,  (c) 
impossible?     (1)  a  =  10,  B  =  50°,  C  =  75°;  (2)  a  =  75,  B  = 
130°,  C  =  65°;  (3)  b  =  12,  A  =  170°,  C  =  10°. 

4.  If  the  given  side  of  a  triangle  is  2",  and  the  two  angles  including 
the  side  are  supplementary,  what  is  the  value  of  the  third  angle 
of  the  triangle? 

102.  Experiment  VIII. — To  draw  a  triangle  of  which  one 
side  and  two  angles  are  given,  either  angle  being  adjacent 
to  the  given  side  and  the  other  angle  opposite  the  given  side. 

Helps. — (1)  Find  the  third  angle  of  the  triangle;  (2)  pro- 
ceed as  in  101.  Notice  that  two  possible  arrangements  of 
the  given  parts  may  be  specified,  but  only  one  triangle 
can  be  drawn  for  each  arrangement. 

EXERCISES 

1.  Construct  triangles  from  the  following  measurements:  (1)  a  = 
3.4",  B  =  72°,  A  =  85°;  (2)  a  =  3.4",  A  =  72°,  B  =  85°;  (3) 
b  =  1.5",  B  =  87°,  C  =  135°. 

2.  Construct  a  right  triangle  with  the  hypotenuse  =  3.5"  and  an 
acute  angle  =  37°. 


TRIANGLES  57 

103.  EXPERIMENT  IX. — To  draw  a  triangle  of  which  the 
three  angles  are  given. 

Helps. — (1)  What  relation  must  exist 
between  the  three  given  angles?  (2) 
Observe  the  number  of  triangles  that 
can  be  constructed  with  the  same  given 
angles. 

EXERCISES 

1.  Show  that  when  three  angles  of  a  triangle  are  known,  no  more 
is  given  than  when  two  angles  of  the  triangle  are  known. 

2.  Construct  an  equiangular  triangle. 

104.  Experiment  X. — Properties  of  an  isosceles  triangle. 
Draw  an  isosceles  triangle.     Observe: 

(1)  The  relative  size  of  the  base  angles. 

Bisect  the  vertex  angle  of  the  triangle.     Observe: 

(2)  Where  the  bisector  meets  the  base. 

(3)  The  position  of  the  bisector  with  respect  to  the  base. 

(4)  The  symmetry  of  the  triangle. 

Draw  and  bisect  an  exterior  angle  at  the  vertex.  Ob- 
serve : 

(5)  The  position  of  the  bisector. 

Extend  one  of  the  equal  sides  of  the  triangle  through  the 
vertex  by  a  distance  equal  to  the  length  of  a  side;  join 
the  extremity  of  the  part  extended  with  the  nearer  end 
of  the  base.  Observe: 

(6)  The  position  of  the  connecting  line. 

State  results,  including  any  other  properties  of  the 
triangle  which  you  discover. 

105.  EXPERIMENT  XI. — To  draw  a  perpendicular  to  a 
given  line  through  a  given  point,  (a)  when  the  point  is  in 
the  line,  and  (b)  when  the  point  is  without  the  line. 

Helps. — Draw  a  straight  line  and  mark  the  given  point. 
Adapt  the  figure  of  104,  part  (6),  to  both  -constructions. 
Thus  in  (6) ,  the  given  line  contains  the  base  of  the  isosceles 
triangle,  and  the  given  point  is  at  the  extremity  of  the 
extended  side  of  the  triangle. 


58  PLANE    GEOMETRY 

106.  EXPERIMENT  XII. — To  draw  a  line  parallel  to  a  given 
line  and  passing  through  a  given  point. 

Helps. — Draw  a  straight  line  and  mark  the  given  point. 
Adapt  the  figure  of  104,  part  (5).  Describe  method. 

107.  Experiment    XIII.— Properties    of    an    equilateral 
triangle. 

108.  Experiment  XIV. — A  triangle  with  two  equal  angles. 
Helps. — Draw  a  side  of  the  triangle;  and  make  the  two 

angles  adjacent  to  the  side,  equal.     Observe  the  kind  of 
triangle  that  is  formed. 

109.  Experiment  XV.— A  triangle   with  three  unequal 
sides. 

Helps. — Draw  the  triangle.  Observe  the  size  of  the 
angles  and  how  they  are  situated  with  respect  to  the  sides. 

EXERCISES 

1.  The  sides  of  a  triangle  are  a  =  5,  b  =  7,  c  =  5.     Into  what 
parts,  or  segments,  does  the  bisector  of  angle  B  divide  side  6? 

2.  The  angles  of  a  triangle  are  70°,  55°,  55°.     What  kind  of  triangle 
is  it? 

3.  The  sides  of  a  triangle  are  a  =  5,  6  =  6,  c  =  7;  the  angles  are 
about  44°,   78°,   58°.     How  are  these  angles  situated  in  the 
triangle? 

110.  Experiment  XVI. — A  line  joining  the  middle  points 
of  two  sides  of  a  triangle. 

Helps. — Draw  the  triangle,  mark  the  middle  points  of 
any  two  sides;  join  these  points  with  a  line.  Observe  (a) 
the  position  of  this  line,  and  (6)  its  length  as  compared 
with  a  side  of  the  triangle. 

111.  Experiment  XVII. — Concurrent  lines  of  a  triangle. 
Draw   acute,    right   and   obtuse  scalene  triangles.     In 

each  triangle,  draw: 

(a)  The  three  altitudes. 
(6)  The  three  medians. 

(c)  The  three  angle  bisectors. 

(d)  The  three  perpendicular-bisectors  of  the  sides. 

(e)  The  bisectors  of  the  three  pairs  of  exterior  angles. 


TRIANGLES  59 

• 

Observe. — (a)  The  concurrence  of  the  altitudes,  the 
orthocenter;  (6)  the  concurrence  of  the  medians,  and  where 
on  each  median  the  point  of  concurrence  is  situated,  the 
centroid]  (c)  the  concurrence  of  the  angle  bisectors,  and  the 
special  position  of  the  point  of  concurrence  with  respect  to 
the  sides  of  the  triangle,  the  incenter;  (d)  the  concurrence 
of  the  perpendicular-bisectors  of  the  sides,  and  the  special 
position  of  the  point  of  concurrence  with  respect  to  the 
vertices  of  the  triangle,  the  circumcenter;  (e)  the  concurrence 
of  each  two  exterior  angle  bisectors  with  the  bisector  of  the 
non-adjacent  interior  angle  (three  points),  and  the  special 
position  of  these  points  of  concurrence  with  respect  to  the 
sides  and  extended  sides  of  the  triangle,  the  excenters. 

112.  EXPERIMENT  XVIII.— To  find  the  center  of  gravity 
of  a  triangle. 

Help. — This  point  is  found  as  in  111,  part  (b). 

113.  Experiment  XIX.— Circles  related  to  a  triangle. 

(a)  The  inscribed  circle,  tangent  to  the  three  sides. 

(b)  The  circumscribed  circle,  containing  the  three  vertices. 

(c)  The  three  escribed  circles,  tangent  to  one  side  and 
to  two  extended  sides. 

Help. — The  centers  of  these  circles  are  found  as  in  111, 
parts  (c),  (d)  and  (e). 

EXERCISES 

1.  Draw  an  isosceles  triangle.     Draw  the  altitudes,  medians,  angle- 
bisectors,  perpendicular-bisectors  of  the  sides,  and  bisectors  of 
the  exterior  angles.     Draw  the  inscribed,  circumscribed  and 
escribed  circles. 

2.  Draw  as  in  Exercise  1,  for  an  equilateral  triangle. 

114.  REVIEW  EXERCISES 

Construct  the  following  triangles  from  the  given  parts: 

1.  A  right  isosceles  triangle  whose  equal  sides  are  given;  that  is, 
equal  to  a  given  sect  or  to  a  given  measurement. 

2.  An  isosceles  triangle  whose  base  and  equal  sides  are  given. 

3.  An  isosceles  triangle  whose  equal  sides  and  base  angles  are 
given. 

4.  An  isosceles  triangle  whose  base  and  vertex  angle  are  given. 


60  PLANE    GEOMETRY 

5.  A  triangle  whose  altitude  and  base  angles  are  given. 

6.  A  triangle  whose  altitude,  base,  and  one  base  angle  are  given. 

7.  An  equilateral  triangle,  whose  altitude  is  given. 

8.  An  isosceles  triangle  whose  base  and  altitude  are  given. 

9.  A  triangle  of  which  one  side,  the  opposite  angle  and  the  alti- 
tude to  another  side  are  given. 

10.  A  triangle  of  which  a  side,  the  altitude  to  the  side  and  the 
median  to  the  side  are  given. 

11.  (a)  A  triangle  of  which  two  sides  are  3.0  and  2.3.  inches,  and 
the  altitude  to  the  shorter  side  is  1.5  in.     (6)  A  triangle  with 
the  same  given  sides  and  the  altitude  to  the  longer  side  1.5  in. 

12.  A  triangle  of  which  a  side,  the  altitude  to  that  side,  and  the 
radius  of  the  circumscribed  circle  are  given. 

13.  Are  any  of  the  constructions  of  examples  1  to  12  impossible 
for  certain  relative  values  of  the  given  parts? 

14.  Construct  accurately  without  a  protractor: 

(a)  an  angle  of    60°;  (d)  an  angle  of    15°; 

(b)  an  angle  of    30°;  (e)   an  angle  of    75°; 

(c)  an  angle  of  120°;  (/)    an  angle  of  105°. 

15.  Draw  a  straight  line  through  a  given  point,  such  that  the  per- 
pendiculars to  the  line  from  two  other  given  points  shall  be 
equal. 

16.  Draw  an  angle  ABC  and  a 
point  P  within  it.     Find  a 
line  DE  passing  through  P, 
such   that  the  part  of  the 
line  included  by  the  sides 
of  the  angle  is  bisected  at 

point  P.     Help. — Apply  a  principle  similar  to  that  of  110. 

17.  Draw  two  sects  a  and  b,  and  two  unequal  angles,  C  and  C',  of 
which  C>C".     Construct  a  triangle  having  two  sides  and  the 
included  angle  equal  respectively  to  a,  b,  C.     Construct  a  second 
triangle  having  two  sides  and  the  included  angle  equal  respec- 
tively to  a,  b,  C'.     Observe  the  relative  lengths  of  the  third 
sides  c  and  c'  of  the  two  triangles. 

18.  Draw  four  sects  a,  b,  c,  c';  of  which  c>c'.     Construct  a  triangle 
having  three  sides  equal  respectively  to  a,  b,  c.     Construct  a 
second  triangle  having  three  sides  equal  respectively  to  a,  b,  c'. 
Observe  the  relative  size  of  the  angles  C  and  C'. 

19.  If  a  triangle  is  drawn  from  the  measurements,  a  =  7,  c  =  12, 
B  =  75°,  and  a  second  triangle  is  drawn  from  the  measurements 
a'  =  7,  c'  =  12,  B'  =  105°,  how  will  the  sides  6  and  6'  of  the 
two  triangles  compare  in  length? 


TRIANGLES  61 

APPLICATIONS 

115.  SURVEYING  AND  ENGINEERING  MEASUREMENTS. — In 
each  of  the  following  problems  make  a  sketch  of  the  figure 
with  the  proper  lettering,  and  write  all  given  values  on 
this  sketch.  Then  select  a  suitable  scale  and  construct  the 
figure  accurately  by  the  methods  of  97-102.  Find  the 
required  values  by  measurement  from  the  completed  figure. 

Accurate  solutions  of  these  problems  may  be  made  by 
trigonometry,  which  is  a  development  of  geometry  and 
algebra  for  the  purpose  of  calculating  the  required  parts  of 
triangles  from  the  given  parts.  A  surveyor  or  engineer 
would  construct  the  figures  to  scale,  as  is  required  in  these 
exercises,  as  a  check  on  the  more  accurate  calculated  values. 

The  line  measurements  given  are  made  with  a  steel  tape 
and  the  angle  measurements  are  made  with  a  transit. 

EXERCISES 

1.  A  tree,  C,  is  observed  from  two  points,  A  and  B,  1000  feet  apart 
on  a  straight  road.     The  angles  CAB  and  CBA  are  respectively 
70°  and  45°.     How  far  is  the  tree  from  each  point?     Use  a 
linear  scale  either  of  500  feet  to  the  inch,  or  of  200  feet  to  the 
inch. 

2.  A  farm  is  bounded  by  three  straight  roads,  the  sides  of  the 
farm  being  850  feet,  1000  feet,  960  feet.     What  angles  do  the 
sides  of  the  farm  make  at  the  three  vertices? 

3.  A  vertical  pole,  22  feet  high,  casts  a  shadow  at  mid-day  35  feet 
long  on  a  horizontal  plane.     What  is  the  angle  of  elevation  of 
the  sun? 

4.  An  aeroplane  is  directly  over  a  field  that  is  known  to  be  3  \  miles 
distant.     The  angle  of  elevation  at  the  point  of  observation  is 
10°.     How  high  above  the  ground  is  the  aeroplane? 

5.  The  distance,  AB,  across  a  pond  is  desired.     A  point  C,  is 
selected  from  which  CA  and  CB  can  be  measured  1300  ft.  and 
890  ft.  respectively,  and  at  which  the  angle  ACB  is  measured 
42°.     Find  AB, 

6.  The  distance  of  a  lighthouse  A,  from  a  point  B,  on  shore  is 
required.     A  line  BC  is  laid  off  on  shore  1500  ft.  long.     The 
angles  between  this  line  and  the  direction  of  the  lighthouse  are 
measured  at  points  B  and  C,  73°  and  81°  respectivley.     How 
far  is  the  lighthouse  from  B? 


62  PLANE    GEOMETRY 

7.  Two  steamers  leave  the  same  port  at  the  same  time.     One  sails 
northwest  15  miles  an  hour;  the  other  east  19  miles  an  hour. 
How  far  apart  are  they  after  5  hours? 

8.  A  bridge  is  to  span  a  river.     The  positions  of  the  abutments 
on  the  banks  of  the  river  are  marked  by  stakes  A  and  B.     A 
stake  C  is  placed  on  the  same  side  of  the  river  as  A,  2460  feet 
along  the  bank;  the  angles  BAG  and  BCA  are  measured  80° 
and  57°  respectively.     Find  distance  AB  across  the  river. 

9.  The  height  of  a  flagpole  is  desired.     At  a  distance  of  130  feet 
from  the  base  the  angle  of  elevation  of  its  top  is  50°.     Find 
the  height  of  the  pole. 

10.  The  distance  between  two  buildings  is  1500  feet,  and  the  dis- 
tance of  a  point  of  observation  from  the  nearer  building  is 
2400  feet.     The  angle  between  lines  from  the  point  of  obser- 
vation to  the  buildings  is  20°.     How  far  is  the  point  of  observa- 
tion from  the  other  building? 

The  omission  of  what  word 
would  make  this  construc- 
tion ambiguous? 

11.  A  surveyor  often  uses  methods 
involving  several  triangles,  in 
order    to   find   some   distance 
that  cannot  be  directly  mea- 
sured.      A    and    B    are    two 

f*  n 

buoys  in  a  harbor,  whose  loca- 
tions are  to  be  marked  on  a  chart.  A  line  CD  is  laid  out  on 
shore,  and  is  also  drawn  to  scale  on  the  chart;  C  being  nearer  to 
A,  and  D  nearer  to  B.  CD  is  measured  500  feet;  angles  ACD, 
BCD,  BDC,  ADC,  are  measured  respectively  115°,  45°,  110°, 
60°.  Plot  accurately  and  measure  AB. 


CHAPTER  V 
TRIANGLES 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

116.  The  First  Step  in  Classification. — The  most  elemen- 
tary principles  relating  to  triangles,  based  upon  the  experi- 
mental results  of  Chapter  IV,  are: 

Postulate  15. — Two  triangles  are  congruent  if  the  three 
sides  of  one  triangle,  are  equal  respectively  to  the  three 
sides  of  the  other  triangle  (97) . 

Postulate  16. — Two  triangles  are  congruent  if  two  sides 
and  the  included  angle  of  one  triangle,  are  equal  respectively 
to  the  two  sides  and  included  angle  of  the  other  triangle 
(98). 

Postulate  17. — Two  triangles  are  congruent  if  two  sides 
and  the  angle  opposite  the  greater  of  the  two  sides  of  one 
triangle,  are  equal  respectively  to  two  sides  and  the  angle 
opposite  the  greater  of  the  two  sides  of  the  other  triangle 
(99). 

Postulate  18. — Two  triangles  are  not  necessarily  congruent 
if  two  sides  and  the  angle  opposite  the  less  of  the  two  sides 
of  one  triangle,  are  equal  respectively  to  two  sides  and  the 
angle  opposite  the  less  of  the  two  sides  of  the  other  triangle 
(100). 

Postulate  19. — Two  triangles  are  congruent  if  one  side 
and  any  two  angles  of  one  triangle,  are  equal  respectively 
to  one  side  and  the  two  homologous  angles  of  the  other 
triangle  (101, 102). 

Postulate  20. — Two  triangles  are  not  necessarily  congru- 
ent if  the  three  angles  of  one  triangle  are  equal  respectively 
to  the  three  angles  of  the  other  triangle  (103). 

These  postulates  mean  that  if  two  triangles  are  known  to 

63 


64  PLANE    GEOMETRY 

have  any  three  parts  respectively  equal  (with  two  excep- 
tions), they  are  alike  in  all  respects;  that  is,  the  remaining 
parts  are  also  respectively  equal. 

117.  The  Second  Step  in  Classification.— Most  of  the  other 
principles  relating  to  triangles  can  be  explained  from  the 
elementary  principles,  or  Postulates,  of  116,  and  from  the 
previously  analyzed  and  classified  principles  of  Chapter  III. 

118.  Theorem  I   (of  triangles). — The  sum  of  two  sides 
of  a  triangle  is  greater  than  the  third  side. 

Help. — Apply  Postulate  4. 

Corollary. — The  difference  between  any   two   sides  of  a 
triangle  is  less  than  the  third  side. 

Help. — Apply  Axiom  3  to  the  conclusion  of  Theorem  I. 

119.  Theorem  II. — The  sum  of  the  angles  of  a  triangle  is 
a  straight  angle  (94). 


State  the  hypothesis  and  conclusion. 

Analysis.— (1)  Extend  AB  to  D  and  draw  BE\\AC] 
what  postulate? — (2)  x  and  y  are  equal  to  angles  of  the 
triangle;  what  postulate  and  theorem  of  parallels? — (3)  x+y 
+  B  =  a  straight  Z;  what  axiom? — (4)  substitute. 

Write  the  preceding  analysis  in  the  form  of  69,  72,  etc. 
This  is  one  of  the  most  famous  theorems,  and  is  believed 
to  have  been  analyzed  by  Pythagoras  and  Aristotle,  as  well 
as  by  Euclid. 

Corollaries. — 1.  A  triangle  has  at  least  two  acute  angles 
(94). 

2.  When  two  angles  of  a  triangle  are  known,  the  third 
angle  can  be  found. 

3.  The  two  acute  angles  of  a  right  triangle  are  complemen- 
tary. 

4.  When  two  angles  of  one  triangle  are  equal  respectively 
to  two  angles  of  another  triangle,  the  third  angles  are  also 
respectively  equal. 


TRIANGLES 


65 


120.  Theorem  III. — An  exterior  angle  of  a  triangle  equals 
the  sum  of  the  two  non-adjacent  interior  angles  (94) . 

Corollary. — An  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  non-adjacent  interior  angles. 

121.  Theorem  IV. — The  base  angles  of  an  isosceles  triangle 
are  equal  (104). 

Hypothesis. — Isosceles  triangle  ABC. 

Conclusion. — B  =  C. 

Analysis. 

STATEMENT  REASON 

1.  Draw  AD  bisecting   /.A     Postulate  12 

2.  Compare  triangles  ABD 
and  A  CD 

AB  =  AC 


3. 

4.  AD  =  AD 

5.  Zz  =    £y 

6.  AABD  * 

7.  :.B  =  C 


ACD 


Hypothesis 

Identical 

Made  so  in  drawing  AD 

Postulate  16 

Homologous  parts  of  ^  A 


The  method  of  analysis  employed,  is  to  draw  an  aux- 
iliary line  which  will  form  two  triangles  with  other  lines  of 
the  figure.  Three  homologous  parts  of  these  two  triangles 
are  known  to  be  equal  from  the  method  of  drawing  the 
figure.  Therefore  the  triangles  are  congruent,  or  alike  in 
all  respects  (with  exceptions) ;  and  therefore  any  remaining 
homologous  parts  of  the  two  triangles  are  equal  (116). 
This  method  of  analysis  is  much  used.  It  will  be  found 
an  aid  in  understanding  congruence  of  triangles,  if  equal 
parts  of  the  two  triangles  under  consideration  are  overlaid 
with  colored  crayons.  Thus  AB  and  AC  may  be  overlaid 
with  a  red  crayon,  AD  for  each  triangle  with  a  blue  crayon, 
and  angles  x  and  y  with  a  yellow  crayon. 

Corollaries. — 1.  An  equilateral  triangle  is  also  equiangular 
(107). 

Help. — Such  a  triangle  is  isosceles  in  several  ways. 

2.  Each  angle  of  an  equilateral  triangle  is  two-thirds  of  a 
right  angle,  or  60°  (107). 


66 


PLANE    GEOMETRY 


122.  Theorem  V. — The  bisector  of  the  vertex  angle  of  an 
isosceles  triangle  bisects  the  base  (104). 

Helps. — Show  that  the  two  triangles  formed  in  the  figure 
are  congruent  by  Postulate  16;  and  therefore  two  homolo- 
gous sides  of  these  triangles  are  equal. 

123.  Theorem  VI. — The  bisector  of  the  vertex  angle  of 
an  isosceles  triangle  is  perpendicular  to  the  base  (104). 

Helps. — Use  the  method  of  121  and  122.  Certain  angles 
of  the  two  triangles  are  therefore  equal.  Then  apply  84, 
Theorem  (9);  or  complete  as  that  theorem  is  analyzed. 

124.  Theorem  VII. — The  median  drawn  to  the  base  of  an 
isosceles  triangle  (a)  bisects  the  vertex  angle  and  (b)  is  per- 
pendicular to  the  base  (104). 

125.  Theorem  VIII. — The  altitude  drawn  to  the  base  of  an 
isosceles  triangle  (a)  bisects  the  vertex  angle,  and  (b)  bisects 
the  base  (104).  E 

126.  Theorem  IX. — The  perpendic- 
ular-bisector of  the  base  of  an  isosceles 
triangle  (a)  contains  the  vertex  of  the 
triangle,  and  (b)  bisects  the  vertex  angle 
(104). 

Hypothesis. — Isosceles  A  ABC,  with 
DE  the  perpendicular-bisector  of  BC. 

Conclusion. — (a)  DE  contains  point 
A;  (b)  DE  bisects  ZA. 


B 


Analysis. 
Part  (a) : 

STATEMENT 

1.  As  DE  may  not  contain 
point  A,  draw  AD 
AD.LBC 
But  DEA.BC 
.'.  DE  coincides  with  AD 
5.  .'.DE  contains  point  A 
127.  A  Postulate.     Postulate  21. — If  two  straight  lines 
coincide,  any  property  of  one  of  the  lines  belongs  to  the 
other  line  also. 


2. 
3. 

4. 


REASON 
Postulate  12 

Theorem  VII  (124) 
Hypothesis 
Postulate  10 
Postulate  21  (see  127) 


TRIANGLES 


67 


128.  A    General    Theorem    of    Isosceles    Triangles.— 
Theorems  V  to  IX  may  be  summarized  thus :  In  an  isosceles 
triangle,  the  bisector  of  the  vertex  angle,  median  to  the 
base,  altitude  to  the  base,  and  perpendicular-bisector  of 
the  base,  all  coincide  in  one  line. 

Corollary. — State  a  similar  theorem  for  an  equilateral 
triangle. 

129.  Theorem  X. — //  two  sides  of  a  triangle  are  unequal, 
the  angles  opposite  the  sides  are  un- 
equal, the  greater  angle  being  opposite 

the  greater  of  the  two  sides  (109). 
Hypothesis. — A  ABC  with  a>b. 
Conclusion. — A  >B. 


A  «* 


Analysis. 

STATEMENT 

1.  Measure  CD  =  AC 
and  draw  AD 

2.  x  =  y 

3.  A>y 

4.  A  >x 

5.  x>B 

6.  A>B 


REASON 
What  postulate? 

What  theorem? 
Axiom  12 
Axiom  13 
What  theorem? 
What  axiom? 


Corollary.  —  //  the  three  sides  of  a  triangle  are  all  unequal 
the  three  angles  are  all  unequal,  the  size  of  the  angles  being 
in  the  same  order  as  that  of  the  sides  to  which  they  are  opposite 
(109). 

130.  Theorem  XI.  —  //  two  angles  of  a  triangle  are  equal, 
the  sides  opposite  these  angles  are  equal  and  the  triangle  is 
isosceles  (108). 

Helps.  —  Draw  the  triangle  with  two  equal  angles.  Draw 
an  auxiliary  line  bisecting  the  third  angle.  Show  that  the 
two  triangles  formed  have  three  parts  respectively  equal 
and  are  congruent;  etc. 

Corollary.  —  An  equiangular  triangle  is  also  equilateral. 

EXERCISE 

Of  what  theorem  is  Theorem  XI  the  converse? 


68  PLANE    GEOMETRY 

131.  Additional  Axioms. — Just  as  the  statement  of  a 
new  postulate  was  found  necessary  in  126  (stated  in  127), 
so  some  new  axioms  will  be  found  necessary  in  Theorem 
XII  (133),  as  there  analyzed. 

EXERCISES 

1.  If  there  are  two  possible  places  of  concealment  of  an  object,  and 
one  of  these  places  has  been  examined  without  finding  the  object, 
where  must  the  object  be? 

2.  If  there  are  three  possible  places  of  concealment  of  an  object,  and 
two  of  them  have  been  examined  without  finding  the  object, 
where  must  it  be? 

3.  How  many  possible  conditions  may  exist  as  to  the  relative  size 
of  two  objects,  A  and  B?     If,  of  these  three  possible  conditions, 
two  have  been  shown  to  be  impossible,  what  may  we  conclude? 

This  elementary  reasoning  leads  to  the  statement  of 
the  following  axioms: 

Axiom  14. — One  of  two  quantities  of  the  same  kind  may 
be  larger  than,  equal  to,  or  smaller  than  the  other. 

Axiom  15. — If,  of  three  possible  relations  of  size  between 
two  quantities  of  the  same  kind,  two  cannot  exist,  then  the 
third  relation  exists. 

132.  Indirect  Analysis. — An  analysis  in  which  a  principle 
is  established  by  disproving  all  other  possible  conditions, 
is  called  indirect.     This  method  is  applied  for  the  first 
time  in  Theorem  XII  (133). 

133.  Theorem  XII. — //  two  angles  of  a  triangle  are  un- 
equal, the  sides  opposite  these  angles  are  unequal,  the  larger 
side  being  opposite  the  larger  angle. 

Hypothesis.— AABC  with  B>C. 

Conclusion. — b  >  c. 

Analysis. 

STATEMENT  REASON 

1.  Three  relations  of  size  are  pos-  Axiom  14 

sible  between  b  and  c. 
(a)  b>c 
(b)b  =  c 
(c)  b<c 


TRIANGLES  69 

2.  Relation   (b),  b  =  c,  is  impos-  Theorem  IV 
sible  since  then  B  would  =  C 

3.  Relation  (c),  b<c,  is  impossible 

Since  then  B  would  be  <  C  Theorem  X 

4.  .'.  Relation  (a)  is  the  true  one,  Axiom  15 
or  b>c 

Corollary. — State  a  corollary  for  three  unequal  angles, 
similar  to  the  corollary  of  Theorem  X. 

EXERCISE 

1.  Of  what  theorem  is  Theorem  XII  the  converse? 

• 

134.  CONVERSE  THEOREMS  OF  ISOSCELES  TRIANGLES. — A 
triangle  is  isosceles,  in  which: 

(1)  The  bisector  of  an  angle  is  perpendicular  to  the  opposite 
side. 

(2)  The  bisector  of  an  angle  bisects  the  opposite  side. 

(3)  A  median  bisects  the  angle  from  which  it  is  drawn. 

(4)  A  median  is  perpendicular  to  the  side  to  which  it  is 
drawn. 

(5)  The  perpendicular-bisector  of  a  side  contains  the  opposite 
vertex. 

Help. — All  of  these  theorems  may  be  analyzed  by  showing 
that  the  two  triangles  into  which  the  given  triangle  is 
divided,  are  congruent;  therefore,  etc. 

135.  The  Position  Assigned  a  Principle  in  the  System 
of    Analysis    and    Classification. — All    the    principles    of 
Chapter  IV  which   were   discovered   and   proved   experi- 
mentally, have  now  been  analyzed  and  classified  in  this 
chapter,  except  the  principles  of  110-113.     These  princi- 
ples cannot  be  analyzed  until  the  logical  system  has  been 
carried  to  a  more  advanced  stage.     The  system  of  classifica- 
tion has,  however,  now  been  developed  to  the  point  where 
certain  principles  of  Chapter  II,  28-32  inclusive,  can  be 
analyzed. 


70 


PLANE    GEOMETRY 


136.  Theorem  XIII. — The  rule  and  compass  construction 
for  bisecting  an  angle  is  correct  (28). 

C 


Hypothesis.—  /.ABC,  BD  =  BE,  DF  =  EF,  BF  drawn. 
Conclusion. — BF  bisects    Z.ABC;  that  is  x  =  y. 
Analysis. 


STATEMENT 

1.  Compare  &BEF  and 
BDF 

2.  BE  =  BD 

3.  EF  =  DF 

4.  BF  =  BF 

5.  A  are  ^ 

6.  .*.  x  =  y 


REASON 


Hypothesis 

Hypothesis 

Identical 

Postulate  15 

Homologous  parts  of  ^  A 


137.  Theorem  XIV. — The  rule  and  compass  construction 
for  bisecting  a  sect  is  correct  (29) . 

Helps.— (I)  Prove  ACAD  * 
ACBD-,  (2)  .'.  x  =  y;  (3)  ACAB 
is  isos.;  (4)  .'.  CE  bisects  AB. 

EXERCISES 

1.  Draw  the  two  triangles  separately 
whose   congruence  is  first   proved. 
Overlay  the  equal  parts  with  colored 
crayons.      Draw   the   isosceles   tri- 
angle which  is  next  considered,  with 
the  bisector  of  its  vertex  angle. 

2.  Why  cannot  &CAE  and  CBE  be  proved  congruent  at  once  from 
the  three  respectively  equal  parts,  CA  =  CB,  CE  =  CE,  s  =  tt 

3.  After  finding  that  x  =  y,  because  the  A  CAD  and  CBD  are  con- 
gruent, can  the  &CAE  and  CBE  then  be  proved  also  congruent? 
Would  this  method  of  completing  the  analysis  or  that  suggested 
be  better? 


TRIANGLES  71 

138.  Theorem  XV. —  The  rule  and  compass  construction 
for  erecting  a  perpendicular  to  a  straight  line  at  a  point  in 
the  line  is  correct  (30). 

Helps. — Use  congruent  triangles  and  84,    Theorem  (9.) 

139.  Theorem  XVI. — State  a  theorem  as  indicated  in 
31. 

Help. — Analyze  by  a  method  similar  to  that  of  137. 

140.  Theorem  XVII. — State  the  theorem  indicated  in  32. 
Analyze  it. 

141.  VARIATIONS  IN  THE  ANALYSIS  OF  THEOREMS. — (1) 
Analyze  Theorem  II  (119)  by  drawing  an  auxiliary  line 
through  C  ||  AB.     This  was  probably  the  method  employed 
by  Pythagoras. 

(2)  Analyze  Theorem  II  (119)  by  drawing  lines  parallel 
to  AC  and  BC  from  any  point  in  AB. 

(3)  Analyze  Theorem  III  (120)  by  employing  the  two 
principles:  the  sum  of  the  angles  of  a  triangle  is  a  straight 
angle;  and,  an  exterior  angle  and  adjacent  interior  angle  are 
supplementary. 

(4)  Analyze  Theorem  VIII  of  Chapter  III  (81)  by  draw- 
ing an  auxiliary  line  through  B  and  E.     Two  pairs  of  cor- 
responding angles  are  equal,  and  therefore  angles  B  and  E, 
which  are  the  differences  of  these  angles  are  equal.     Simi- 
larly for  angles  B  and  H. 

(5)  Analyze  Theorem  IX  of  Chapter  III  (82)  by  extending 
lines  BA,  BC  and  EF  until  they  intersect,  thus  forming  two 
right  triangles  containing  respectively  the  angles  B  and  E. 
Relations  between  the  angles  of  these  triangles  establishes 
the  equality  of  angles  B  and  E.  A 

(6)  Analyze  Theorem  XII  (133), 

by   a   direct   method,    as  follows:       c 

Draw    BD    so    that   x  =  C;   then 

BD  +  DA>c:  .'.  b>c.  T~ 

142.  The   Third   Step   in   Classification.— A   system   of 
classification  in  which  all  known  experimental  principles 
have  been  assigned  a  place  in  a  proper  logical  sequence, 
may  be  extended  into  the  realms  of  investigation  and  dis- 


72 


PLANE    GEOMETRY 


covery.  Many  principles  are  less  evident  than  those 
which  form  the  framework  of  the  system,  and  would 
probably  never  have  become  known  without  the  develop- 
ment of  the  analytic,  or  logical  method. 

Some  of  the  principles  stated  in  143  have  probably  been 
thus  discovered;  as  Theorems  (8),  (14),  (15),  (16),  (17),  (18), 
(31),  (42),  and  (47)  to  (53.) 

ADDITIONAL  THEOREMS 

143.   (1)  The  angle  formed  by  two  lines  drawn  from  any 
point  within  a  triangle  to  two  vertices 
of  the  triangle,  is  greater  than  the 
third  angle  of  the  triangle. 

Helps.—  Z1>Z2;     Z2>Z3;    /. 
Z1>Z3. 
Give  reasons. 

(2)  The  bisectors  of  any  two  angles 
of  a  triangle  intersect. 

Helps. — Show  that  the  angle  formed  by  the  bisectors  is 
(a)  less  than  a  straight  angle,  and  (6)  greater  than  the  third 
angle  of  the  triangle.  (The  preceding  theorem.) 

(3)  Two   lines   drawn   perpendicular   respectively   to   two 
intersecting  lines,  also  intersect. 

Helps.— Zsl  +  2>0°;  .'•  Zs3 
+  4<180°;  .'.  Z5>0°;  .'.  DE 
and  FG  intersect. 

(4)  A  straight  line  is  terminated 
by  two  parallel  lines;  through  its 
middle  point  another  straight  line 
is  drawn  also  terminated  by  the 
parallels.       The    second    line   is 
bisected  by  the  first  line. 

(5)  Two  sects,  AB  and  CD,  are 
drawn  bisecting  each  other.    Then 

AC  and  BD  are  equal  and  parallel.  State  a  similar  property 
of  two  other  lines  that  may  be  drawn  in  the  figure. 

Helps. — (1)  Use  congruent  triangles  to  show  that  AC  = 


TRIANGLES  73 

BD;  (2)  a  pair  of  homologous  angles  of  these  triangles  are 
equal,  and  therefore  the  lines  are  parallel  by  72. 

(6)  //  one  of  the  equal  sides  of  an  isosceles  triangle  is  pro- 
duced through  the  vertex  by  a  distance  equal  to  the  length  of  the 
side,  and  the  extremity  of  the  produced  sect  is  joined  to  the  nearer 
extremity  of  the  base,  this  line  is  perpendicular  to  the  base. 

Helps. — (1)  Two  isosceles  triangles  are  formed,  having 
equal  angles  (121) ;  (2)  two  of  these  angles  together  form  a 
right  angle. 

Where  has  this  principle  been  established  experimentally? 

(7)  Two    chords    of   a    circle    which  are 
drawn  from  the  same  point  on  the  circum- 
ference  and  which  pass  through  the  extrem- 
ities of  a  diameter,  are  perpendicular. 

Help. —  Identify  the    figure    with 
Theorem  (6). 

(8)  The  shortest  side  of  a  right  triangle  whose  acute  angles 
are  30°  and  60°,  is  one-half  the  hypotenuse. 

Helps. — Draw  an  auxiliary  line  dividing  the  right  angle 
into  two  parts  of  30°  and  60°.  One  of  the  triangles  formed 
is  equilateral  and  the  other  is  isosceles. 

(9)  The  exterior  angles  at  the  base  of  an  isosceles  triangle 
are  equal. 

(10)  The  hypotenuse  of  a  right  triangle  is  the  longest  side. 

(11)  AD±BC,  AB  =  AC  =  AD.     Then 
ADBC  is  a  right  isosceles  triangle. 

(12)  The  bisector  of  the  exterior  angle  at 
the  vertex  of  an  isosceles  triangle  is  parallel 
to  the  base. 

Where  has  this  principle  been  established  experimentally? 

(13)  State  and  analyze  the  converse  of  Theorem  (12). 

(14)  The  exterior  angle  at  the  vertex  of  an  isosceles  triangle 
is  double  either  (interior)  base  angle. 

(15)  State  and  analyze  the  converse  of  Theorem  (14). 

(16)  A B  is  the  diameter  of  a  circle  with  its  center  at  0;  AC 
is  a  chord',  OC  is  a  radius.     Then   ZBOC  =  2  /.BAC. 

Help. — Apply  Theorem  (14). 


74  PLANE    GEOMETRY 

(17)  The  bisectors  of  the  base  angles  of  an 
isosceles  triangle  form  an  angle  equal  to  an  ex- 
terior angle  at  the  base  of  the  triangle. 

Help.— Apply  120. 

(18)  The  angle  formed  by  the  altitudes  drawn  to 
the  equal  sides  of  an  isosceles  triangle,  is  supple- 
mentary to  the  vertex  angle  of  the  triangle. 

(19)  In  a  AABC,  ZA  is  double  LB.     Then  the 
bisector  of  /.A  meets  BC  in  a  point  D  such  that 
AD  =  BD. 

(20)  //  equal  distances  are  measured  in  succession  on  the 
sides   of  a   square,   and  these  points  are  joined, 
another  square  is  formed. 

Helps. — (1)  The  figure  has  equal  sides;  (2)  its 
angles  are  right  angles. 

(21)  One  of  the  equal  sides  of  an  isosceles  triangle  is  greater 
than  one-half  the  base. 

(22)  The  bisectors  of  the  base  angles  of  an  isosceles  triangle 
form  another  isosceles  triangle  with  the  base. 

(23)  The  bisectors  of  the  exterior  angles  at  the  base  of  an 
isosceles  triangle  form  another  isosceles  triangle  with  the  base. 

(24)  AABC    is    isosceles    with    AB  =  AC;    BD  =  CE. 
Then  &BCD  *  ACBE.  A 

(25)  The    medians    drawn    to   the 
equal  sides  of  an  isosceles  triangle  are 
equal. 

Help. — Use  congruent  triangles  (two  ways). 

(26)  The  altitudes  drawn  to  the  equal  sides  of  an  isosceles 
triangle  are  equal. 

(27)  The  bisectors  of  the  base  angles  of  an  isosceles  triangle 
extended  to  meet  the  opposite  sides,  are  equal. 

(28)  State  a  theorem  concerning  the  lengths  of 
the  medians,  altitudes,  and  angle-bisectors  of  an 
equilateral  triangle. 

(29)  A  ABC  is  isosceles  with  AB  =  AC;  the 
sides  are  produced  so  that  AD  =  AE;  BE,  CD  and 
DE  are  drawn.     Then  BE  =  CD,  and  DE\\BC. 


TRIANGLES 


75 


(30)  Homologous  medians,  altitudes  and  angle-bisectors  of 
two  congruent  triangles  are  equal. 

Help. — The  smaller  triangles  formed  within  the  given 
triangles,  and  containing  the  homologous  lines  as  sides, 
may  be  proved  congruent. 

(31)  Triangle    ABC    has    an 
exterior    angle    C'}    at    C.     The 
bisectors  of  interior  angle  B  and 
exterior    angle    Cf,    meet    at   E. 
Angle  E  =  %  angle  A. 

Helps.—  ZC"    =    ZA    +   Z5; 
Zz  =  Z#  +  Z.y;  ZC"  =  2Zz;  etc. 

(32)  Two  isosceles  triangles  are  congruent  if  their  vertex 
angles  and  equal  sides  are  respectively  equal. 

(33)  //  two  sides  of  one  triangle  are  equal  respectively  to  two 
sides  of  another  triangle,  but  the  included  angle  of  the  first 
triangle  is  less  than  the  included  angle  of  the  other  triangle, 
then  the  third  side  of  the  first  \/_ 
triangle  is  less  than  the  third 

side  of  the  other. 


E  d\  F 

Hypothesis— &  ABC    and    DEF    with    a  =  d,    c  =  f, 


Conclusion. — A  C  <  DF. 

Analysis.— (1)  Place  AABC  in  the  position  A'EF;—(2) 
draw  EG  bisecting  £DEA';—  (3)  ADEG  ^  AA'EG;—(4)  . ' . 
GD=GA';—(5)  but  FG  +  GA'>A'F;—($)  .'.  DF>AC. 

(34)  State  the  converse  of  Theorem  (33.) 

Helps. — Use  the  indirect  method  of  analysis  depending 
upon  Axioms  14  and  15. 

(35)  The  symmetrical  impression  of  a  straight  line  with 
respect  to  a  center  of  symmetry  is  parallel  to  the  given  line. 


76 


PLANE    GEOMETRY 


Helps. — (1)  Two  triangles  are  congruent;  (2)  two  angles 
are  equal;  (3)  .'.  the  lines  are  parallel. 

(36)  AB\\DE;    any    trans- 
versal BC;  BD  and  BE  bisecting 
Zs  x  and  y.     Then  CD  =  CE. 

Helps.— (1)  Zl  =  Z2  and 
Z3  =  Z4;— (2)  .'.  &BCD  and 
BCE  are  both  isosceles; — 
(3)  /.  CD  and  CE  are  both  equal  to  BC. 

(37)  Analyze  Theorem  X  (129) 
as  follows: — (1)  Draw  CD  bisect- 
ing  Z  C,  and  make  CE  =  CA;— 
(2)    A  CAD   *   ACDE;—  (3)  .'. 
Z#  =  /.A; — (4)  .'.  it  can  be  shown    A 
that  ZA 


(38)  The  sum  of  two  lines  drawn  from  any  point  within  a 
triangle  to  the  extremities  of  one 

of  the  sides,  is  less  than  the  sum 
of  the  other  two  sides  of  the  tri- 
angle. 

Helps.— (1)    In    A  ACE,   AC 
+  CE>AD  +  DE;—(2)  in   A 
DEB,  DE+EB>DB;—(3)  add  these  inequalities  (Axiom  9) ; 
—(4)  eliminate  DE  (Axiom  7) ;— (5)  the  conclusion  follows. 

(39)  Two  oblique  lines  drawn  from 
the  same  point  in  a  perpendicular  and 
cutting  off  equal  distances  from  the  foot,    A 
are  equal  (50). 

(40)  State  the  converse  of  Theorem 
(39.) 

(41)  A  perpendicular  is  the  shortest  distance  from  a  point 
to  a  straight  line  (50) . 

Help.— Apply  133. 

(42)  The  sum  of  three  lines  drawn  from  any  point  within  a 
triangle  to  the  three  vertices  is  (a)  less  than  the  sum  of  the  sides, 
and  (6)  greater  than  one-half  the  sum  of  the  sides. 


TRIANGLES 


77 


Helps. — (a)  Apply  Theorem  (38)  for  three  positions  of  the 
figure,  and  add  the  inequalities;  (6)  apply  118. 

(43)  In  a  quadrilateral  A  BCD,  AB  is  the  longest  of  the 
four  sides  and  CD  is  the  shortest.     Then  ZC  is  greater  than 
/.A,  and  /.D  is  greater  than  /.B. 

Help. — Draw  a  diagonal  and  apply  129. 

(44)  The  sum  of  the  three  exterior 
angles  of  a  triangle  (one  at  each  vertex) 
is  a  constant  value. 

Helps. — State   the   value    of   each 

exterior   angle   in   terms   of    interior 

angles;  add;  substitute  the  known  value  of  Zs  a  +  b  +  c. 

(45)  //  equal  segments  are  laid  off 
on  a  straight  line  and  parallels  are 
drawn  through  the  points  of  division, 
any  other  line  cutting  the  parallels  is 
also  divided  into  equal  segments. 

(46)  Any  side  of  a  triangle  is  less 
than  one-half  the  perimeter. 

(47)  An  altitude  BD,  and  a  me- 
dian BE,  are  drawn  to  the  hypotenuse 
of  the  right  triangle  ABC.      Then 

/LDBE  =    ZA  --  ZC. 

(48)  The  bisectors  of  the  exterior  angles  at 
B  and  C  of  AABC,  meet  at  D.      ZD  =  90° 


z 


z 


(49)  An  altitude  drawn  to  any  side  of  a 
triangle,  divides  the  angle  from  whose  vertex 
it  is  drawn  into  two  angles  whose  difference 
equals  the  difference  between  the  other  angles 
of  the  triangle. 

(50)  The  perimeter  of  any  quadrilateral  is  greater  than  the 
sum  of  the  diagonals. 

Helps. — Apply  118  to  four  triangles  in 
such  a  way  that  the  sides  of  the  quadrilat- 
eral are  placed  on  the  greater  sides  of  the 
inequalities;  add  and  divide  by  2. 


B 


78  PLANE    GEOMETRY 

(51)  The  sum  of  the  diagonals  of  any  quadrilateral  is  greater 
than  ike  sum  of  either  pair  of  two  opposite  sides. 

Helps. — Apply  118  to  two  of  the  small  triangles  in  such 
a  way  that  the  segments  of  the  diagonals  are  placed  on  the 
greater  side  of  the  inequalities;  add. 

(52)  In  AABC,  AOAB, 
BD  =  CE.      Then  CD>BE. 

Helps.  —  (1)  ^ABC> 
LACB;  why?— (2)  then  apply 
Theorem  (33)  of  this  section, 
to  the  A's  DBC  and  EBC. 

(53)  The  median  drawn  to  a  side  of  a  tri- 
angle is  less  than  one-half  the  sum  of  the  other 
two  sides. 

Helps.— (I)    AE<    what    two   lines?— (2) 
AABD  ^  AECD-,  etc. 

144.  The  Fourth  Step  in  Classification.— 
There  is  always  an  uncertainty  as  to  the 
correctness  of  principles  whose  discovery  has  depended 
entirely  upon  analysis.  For  it  must  be  remembered  that 
they  are  only  an  outgrowth  of  some  of  the  postulates  and 
previously  analyzed  principles,  or  theorems.  If,  however, 
any  errors  exist  in  the  system  of  analysis  and  classification 
up  to  this  point,  they  can  best  be  detected  by  putting  such 
deduced  principles  to  an  experimental  test. 
EXERCISES 

1.  Draw  a  right  triangle  whose  acute  angles  are  30°  and  60°. 
Measure  the  shortest  side  and  the  hypotenuse  and  test  the 
relation  of  143,  Theorem  (8). 

2.  Draw  an  isosceles  triangle,  with  an  exterior  angle  at  the  vertex. 
Measure  this  exterior  angle  and  the  base  angles  of  the  triangle 
with  the  protractor,  and  test  the  principle  of  143,  Theorem  (14). 

3.  Draw  a  triangle  by  constructing  (1)  a  base,   (2)  one  of  the 
angles  adjacent  to  the  base,  (3)  a  side  any  length,  (4)  an  exterior 
angle  double  the  given  base  angle,   (5)  thus  completing  the 
triangle.     Measure  the  sides  and  test  the^jproperty  of  143, 
Theorem  (15). 

4.  Test  the  principle  of  143,  Theorem  (16),  by  constructing  the 
figure  and  measuring  the  angles  with  a  protractor. 


TRIANGLES  79 

5.  Test  the  principle  of  143,  Theorem  (17),  by  constructing  the 
figure  and  measuring  the  angles  with  a  protractor. 

6.  Test  the  principle  of  143,  Theorem  (18),  by  measurement. 

7.  Test  the  principle  of  143,  Theorem  (31),  by  measurement. 

8.  Test  the  principle  of  143,  Theorem  (42),  by  measuring  the  three 
sides  of  the  given  triangle  and  the  three  lines  drawn  from  the 
point  within  it  to  the  vertices.     The  sum  of  the  three  inner 
lines  should  lie  between  two  certain  values,  wherever  their 
point  of  concurrence  is  taken. 

9.  Test  the  principle  of  143,  Theorem  (47). 

10.  Test  the  principle  of  143,  Theorem  (48). 

11.  Test  the  principle  of  143,  Theorem  (49). 

12.  Test  the  principle  of  143,  Theorem  (50). 

13.  Test  the  principle  of  143,  Theorem  (51). 

14.  Test  the  principle  of  143,  Theorem  (53). 

145.   REVIEW  EXERCISES 

1.  One  acute  angle  of  a  right  triangle  is  25°  36'.     What  is  the 
other  acute  angle? 

2.  An  exterior  angle  of  a  triangle  is  98°  and  a  non-adjacent  interior 
angle  is  57°.     Find  the  interior  angle  at  the  third  vertex. 

3.  Calculate  the  third  exterior  angle  of  a  triangle  two  of  whose 
exterior  angles  are  110°  32'  and  75°  41'. 

4.  The  exterior  angle  at  the  vertex  of  an  isosceles  triangle  is  142° 
30'.     Calculate  the  base  angles. 

5.  May  two  sides  of  one  triangle  be  equal  respectively  to  two 
sides  of  another  triangle,  and  still  the  two  triangles  be  not 
congruent? 

6.  May  two  sides  of  a  right  triangle  be  equal  respectively  to  two 
sides  of  another  right  triangle,  and  still  the  two  triangles  be 
not  congruent? 

7.  Does  the  median  of  a  triangle  bisect  its  angle? 

8.  What  is  the  value  of  the  angle  formed  by  the  bisectors  of  the 
acute  angles  of  a  right  isosceles  triangle? 

9.  If  three  sticks  are  pinned  together 
with  a  single  rivet  at  each  intersec- 
tion,   can   the  triangle  formed  be 
changed  in  shape  without  bending 

or  breaking  the  sticks?  Upon  what 
geometric  principle  does  the  rigidity  de- 
pend? 

10.  If  four  sticks  are  pinned  together  in  the 
same  way,  is  the  frame  rigid?  How  can 
another  stick  be  added  to  make  it  rigid? 


80  PLANE    GEOMETRY 

11.  If  any  point  is  taken  within  the  triangle  whose  sides  are  7,  11, 
15  inches,  and  lines  are  drawn  from  the  point  to  the  extremities 
of  the  11  inch  side,  between  what  values  will  the  sum  of  these 
two  lines  be  found? 

12.  If  lines  are  drawn  from  the  point  (Exercise  11)  to  the  three 
vertices  of  the  triangle,  between  what  values  will  the  sum  of 
the  three  lines  be  found?     Will  the  sum  of  the  three  interior 
lines  vary  with  the  position  of  the  point? 

13.  Under  which  of  the  Postulates  of  116  is  each  of  the  following 
principles  of  right  triangles  included?     Two  right  triangles  are 
congruent: 

(a)  If  the  two  perpendicular  sides  of  one  triangle  are  equal  re- 
spectively to  the  two  perpendicular  sides  of  the  other. 

(6)  If  the  hypotenuse  and  one  of  the  perpendicular  sides  of  one 
triangle  are  equal  respectively  to  the  hypotenuse  and  a  per- 
pendicular side  of  the  other. 

(c)  If  the  hypotenuse  and  an  acute  angle  of  one  triangle  are  equal 
respectively,  etc. 

(d)  If  one  of  the  perpendicular  sides  and  the  adjacent  acute 
angle  of  one  triangle  are  equal  respectively,  etc. 

(e)  If  one  of  the  perpendicular  sides  and  the  opposite  acute  angle 
of  one  triangle  are  equal  respectively,  etc. 

14.  Under  which  of  the  Postulates  of  116  is  this  statement  included? 
Two  isosceles  triangles  are  congruent  if  the  two  equal  sides  and 
one  of  the  base  angles  of  the  two  triangles  are  respectively 
equal. 

15.  Why  is  it  sometimes  necessary  to  separate  the  classification  of 
principles  which  evidently  are  closely  associated  and  have  been 
so  considered  when  they  were  investigated  experimentally? 
Give  examples  of  such  principles,  some  of  which  have  been 
analyzed  in  Chapter  III  and  some  in  Chapter  V. 

APPLICATIONS 

146.  Surveying. — Land  surveying  is  largely  responsible 
for  the  development  of  geometric  science,  which  in  its  turn 
has  made  scientific  surveying  possible. 

EXERCISES 

1.  A  surveyor  measured  the  three  angles  of  a  triangular  field, 
37°  40.3',  81°  55.0',  60°  24.5'.     What  check  exists  as  to  the 
accuracy  of  the  measurements? 

2.  It  is  often  necessary  to  find  the  value  of  an  angle  which  cannot 
be  directly  measured.     In  115,  Exercise  6,  calculate  the  value 


TRIANGLES 


81 


B 


of  the  angle  formed  by  the  two  lines  converging  at  the  light- 
house. 

3.  If  the  sides  of  a  triangular  field  have  been  measured  with  a 
tape,  may  a  map  of  the  field  be  drawn  to  scale? 

4.  In  what  different  ways  may  the  least  number  of  parts  of  a 
triangular  field  be  measured  so  that  a  map  of  the  field  may  be 
drawn? 

5.  AB  is  a  known  side  of  a  field  A  BCD, 
etc.;  T  is  a  tree  whose  position  is 
to  be  located  on  the  map.     What 
different  methods  may  be  employed 

to  locate  the  tree,  each  giving  suffi-          A  B 

cient  data  for  the  construction  of  the  triangle  ABT  to  scale? 

6.  A  farm  is  bounded  by  four  straight  lines.     What  measurements 
of  lines  only  may  be  made  in  order  to  construct  a  map  to  scale? 

7.  The  distance  AB  across  a  pond  may  be  found  by  laying  out  on 
the  ground  a  triangle  congruent  to 

the  triangle  of  which  one  side  is  the    A 
required  distance.     Point  C  is  any 
point  from  which  AC  and  BC  can 
be  measured.     Explain  how  the  re- 
maining points  are  located,  what  line 
equals   the   required  line  A B}   and   why    the    triangles    are 
congruent. 

8.  This  method  may  be  used  when  the  required  distance  AB  is 
across  a  river.     At  A  a  right  angle  is  turned 

(see  143,  Theorem  (6),  and  any  distance 
AC  is  measured  on  the  perpendicular.  'An 
equal  distance  CD  is  measured.  At  D 
another  right  angle  is  turned.  The  sur- 
veyor finds  a  place  E  in  the  perpendic- 
ular, from  which  he  sights  C  and  B  in  a 
straight  line,  and  drives  stake  E.  What 
line  is  equal  to  the  required  line  AB? 
What  triangles  are  congruent  and  why? 

9.  A  triangle  is  sometimes  used  in  surveying, 
formed  by  measuring  A  B  along  a  given  line 
=  50  feet,  and  then  holding  a  100-foot  tape 
in  the  position  ACB,  with  C  at  the  50-ft. 
mark.     What  is  the  value  of  angle  A  ? 

10.  A  surveyor  may  extend  a  line  AB  beyond 

an  obstacle,  as  a  building  or  pond,  and  also  determine  the 
unmeasured  interval  BD,  by  the  method  indicated  in  the  figure. 
The  small  triangles  are  constructed  as  in  Exercise  9.  What 


82 


PLANE    GEOMETRY 


measurements  are  made  to  locate  points  C  and  Z>?     Why  is 
line  DE  an  extension  of  AB1     How  is  BD  found? 


11.  A  line  AB  may  also  be  extended  beyond  an  obstacle  by  the 
method  illustrated  in  the  figure.  What  measurements  must 
be  taken  in  addition  to  the  angular  measurements  indicated? 
The  distance  BD  cannot  be  found  by  this  method  except  by 
constructing  the  triangle  BCD  to  scale  on  paper  and  measuring 
the  length  BD  to  the  scale  of  the  drawing. 


The  methods  described  in  Examples  7,  8,  9  and  10  are  not 
employed   in  high-grade   work.     More  accurate  methods  are 
described  in  the  examples  of  116,  the  required  distances  being 
found  by  trigonometric  calculation. 
12.  A  perpendicular  to  a  line 
AB  from  a  point  C  can 
be  laid  out  very  accur- 
ately  as  follows:  Set  any 
point  D  on  line  AB,  or 
use  the  end  point  B  if 
not    too    far    away    and 
visible  from   C.       Set  a 
transit  at  D  and  measure 
/.ADC.     Set  the  transit 
at  C  and  lay  off    Z.DCE 
=  90°  -  ZADCjandset 
a  stake  at  E  on  line  AB. 


TRIANGLES  83 

The  setting  of  a  stake  at  E  may  be  done  accurately  by  using 
two  transits,  one  at  B  or  D  sighted  on  A,  and  the  other  at  C  sighted 
along  CE;  or  as  follows:  When 

transit  is  set  up  at  B  or  D  set  two          H -H -H 

or  more  stakes  in  line  A  B,  as  at  F, 

G,  H.     Stretch  a  string  over  the  tops 

of  the  stakes.     Then  stake  E  is  set  to  this  string,  with  the 

transit  set  up  at  C. 

147.  LIGHT  REFLECTED  FROM  MIRRORS. — The  cases  of  a 
single  mirror,  and  of  two  parallel  mirrors,  were  considered 
in  86.  The  present  problems  involve  principles  of  triangles. 

EXERCISES 

1.  Show  that  if  two  mirrors,  M  and  AT,  are  perpendicular,  the 
incident  ray  A  is  parallel  to  the  finally  reflected  ray  B. 
Help.— Show  that  a  +  0  =  180°. 


2.  Show  that  if  two  mirrors,  M  and  M',  are  set  at  an  angle  of  60° 
with  each  other,  that  the  finally  reflected  ray  B  makes  an  angle 
of  60°  with  ray  A. 


3.  Show  that  if  two  mirrors,  M  and  M' ,  are  set  at  an  angle  of  45° 
with  each  other,  the  reflected  ray  B  makes  a  right  angle  with  the 
incident  ray  A. 


84 


PLANE    GEOMETRY 


Helps.— (1)  2  Z2  +  y  =  180° 

(2)  2  Z3  +  z  =  180° 

(3)  Z2  +  Z3  =  135° 

(4)  z         +  y  +z  =  180° 


M 


Eliminate  Z2  and  Z3  from  Eq.  (1),  (2)  and  (3);  substitute  value  of 
y  +  z  in  Eq.  (4). 

4.  Show  that  if  two  mirrors,  M  and  M'  , 
are  set  at  an  angle  0,  the  angle  a  be- 
tween the  directions  of  the  incident  and 
finally  reflected  rays  is  equal  to  2/3. 

Helps.—  (1)  x  =  x'  and  y  =  y';  (2) 
a  =  2y  -  2x  (120);  (3)  0  =  y  -  x; 
(4)  /.  a  =  2/3. 

Notice  the  application  of  143,  Theorem 
(31),    to     the     triangles     aM'M    and 

&M'M> 

5.  Does    the    result    of   Exercise   4   agree 
with     those    of    the    three    preceding 
exercises? 

148.  INSTRUMENTS  WHICH  MEASURE  ANGLES  BY  THE  USE 
OF  MIRRORS.  The  Optical  Square.  —  This  instrument  (Fig  .1) 
is  used  for  laying  out  a  line  perpendicular  to  a  given  line. 

6 


FIG.  1. 


FIG.  2. 


The  instrument  is  held  at  a  point  A  (Fig.  2  on  line  AB 
so  that  object  B  can  be  seen  through  the  windows  Q.     At 


TRIANGLES 


85 


M 


the  same  time  the  reflection  of  an  object  C  lying  on  the 
perpendicular  AC  is  seen  in  the  mirror  Mf,  apparently 
in  line  with  B.  Upon  which  of  the  mirror  formations 
of  147  does  the  operation  of  this  instrument  depend? 

The  Sextant. — This  instrument  is  used  to  measure  angles 
where  rapidity  is  necessary  rather  than  great  accuracy,  as 
in  army  reconnoitering.  It  is  also  used  on  boats  since  a 
transit  cannot  be  used 
on  account  of  the 
motion  of  the  vessel. 
It  is  the  only  accurate 
angle-measuring  in- 
strument that  can  be 
so  used.  The  instru-  M' 

ment  consists  of  a 
graduated  arc  of  a 
circle  of  about  sixty 
degrees  (hence  its 
name) ,  about  the  cen- 
ter, 0,  of  which  moves 
an  arm  D.  The  arm 
D  is  shown  at  the 
zero  reading  of  the 
arc  (Fig.  3).  In  this 
position  the  mirrors 
M  and  M'  are  parallel. 
The  mirror  M'  is  fixed  to  the  frame,  and  is  so  placed 
that  a  reflection  of  an  object  may  be  seen  in  it  by  looking 
through  telescope  T,  at  the  same  time  that  another  object 
may  be  seen  directly  through  the  unsilvered  part  of  M'. 
The  mirror  M  is  mounted  on  the  movable  arm  D. 

In  order  to  read  an  angle  BCA,  or  a,  (Fig.  3),  the  sextant 
is  held,  either  horizontally  or  vertically  as  the  case  may  be, 
so  that  an  object  on  one  side  of  the  angle  is  seen  directly 
ahead  at  B.  The  arm  D  is  then  moved  to  position  D' 
so  that  the  object  A  on  the  other  side  of  the  angle  is  seen 
through  the  telescope  by  double  reflection  from  mirrors 


FIG.  3. 


86  PLANE    GEOMETRY 

M  and  M' ',  as  if  it  also  were  straight  ahead  and  coincided  in 
position  with  B.  Then  by  147,  Exercise  4,  angle  /3'=i  a. 
Since  0  =  ft',  the  angle  may  be  read  on  the  scale  ss',  and 
a  =  2  arc  sx.  The  scale  ss'  is  divided  into  degrees  of 
one-half  size  and  thus  the  angle  a  is  read  in  its  true  value. 

The  sextant  is  used  on  ship-board  to  measure  the  alti- 
tude angles  of  the  sun,  moon,  planets  or  stars;  that  is,  the 
angle  from  the  horizon  to  the  star  or  to  the  lower  or  upper 
edge  of  the  sun  or  moon.  From  these  angles  the  location 
of  the  vessel  in  latitude  and  longitude  are  calculated. 
Dark  glasses  are  provided  and  must  be  turned  into  position 
when  the, sun  is  observed. 

149.  Navigation. — The  location  of  a  vessel  at  sea  depends 

entirely  upon  geometric 
principles.  The  distance 
travelled  by  a  vessel  is 
measured  directly  by  a  log,  which  is  dragged  in  the  water 
by  a  cord  and  which  registers  the  distance  by  the  revolution 
of  a  propeller. 

Angles  are  measured  with  a  sextant  (148). 

EXERCISES 

1.  The  distance  of  a  ship  from  a  lighthouse  or  other  object  is  deter- 
mined by  the  following  method.  At  any  place  A,  the  angle  x  is 
measured;  the  ship's  course  is  kept  in  a  straight  line  until  a 
position  B  is  reached  at  which  angle  y  is  found  to  equal  twice 
/.x.  The  distance  A  B  is  measured  meanwhile  by  a  log.  Prove 
that  BC  =  AB.  This  theorem  is  contained  in  143. 


\ 
\ 
\ 


\ 
\ 
\ 
\ 
\ 


B  A 

2.  Another  method  of  determining  the  distance  at  which  a  vessel 
passes  an  object  C,  is  to  observe  position  A,  at  which  /.x  —  45°, 
and  position  B  at  which  /.y  =  90°.  What  other  measurement 
is  made?  How  is  BC  found? 


TRIANGLES 


87 


These  methods  are  known  to  navigators  as  "  Doubling 
the  angle  on  the  bow." 

150.  Engineering. — All  engineering  structures  such  as 
cranes,  roof  and  bridge  trusses,  scaffolding,  etc.,  where 
rigidity  is  necessary,  are  framed  in  a  system  of  triangles. 


AAAAA 


B 


EXERCISE 


If  a  single  cross  brace  as  AB,  were  to  be  omitted  in  a  framed 
structure,  would  the  structure  be  rigid?  Could  it  then  collapse 
without  breaking  a  member?  (See  145,  Exercises  9  and  10.) 

151.  BUILDING  AND  CARPENTRY.  —  Geometric  principles 
are  employed  in  all  construction  work. 

EXERCISES 

1.  A  carpenter  uses  the  following  method  to  bisect  an  angle,  ABC. 
Measure  BD  =  BE;  place  a  carpenter's  square  so  that  the 
edges  GF  and  GH  pass  through  points  D  and  E,  and  so  that  the 

•c      " 


scale  readings  from  G  on  the  two  arms  are  equal.  Show  that 
the  method  is  correct.  May  the  square  be  turned  over  so  as  to 
bring  the  vertex  G  nearer  to  point  B? 

2.  A  construction  similar  to  that  of  Exercise  1  may  be  made  by 
using  your  triangle,  by  laying  off  scales  on  the  two  perpendicular 
sides.     Bisect  an  angle  by  this  method. 

3.  A  carpenter  may  use  the  following  method  to  bisect  angle  ABC, 


88 


PLANE    GEOMETRY 


Exercise  1.     Measure  BD  =  BE;  draw  DE;  bisect  Db    meas- 
urement at  K;  draw  BK.     State  the  princi- 
pie  involved. 

4.  The  builder's  plumb  level  is  sometimes  used 
to  set  two  points  at  the  same  level.     Show 
that  B  and  C  are  at  the  same  level  when 
the  plumb  line  hangs  at  the  middle  point 
Z),  of  the  cross-piece. 

and  AB  =  AC. 

5.  A  carpenter  is  sometimes  required  to  cut  two  timbers  x  and  y, 
so  that  a  third  timber  z  shall  make  equal  angles  with  both  of 
them.     Explain  how  he  may  use  the  two  squares  and  why  AB 
is  the  required  line  on  which  to  cut  x  and  y. 


AAEF  is  isosceles, 


6.  A  board  may  be  divided  into  any  number  of  strips  of  equal 
width  by  the  method  shown.  Explain  why  the  perpendicular 
line  AB  is  divided  equally  (143,  Theorem  (45)). 

152.  PROBLEMS  FOR  FIELD  WORK. — Suitable  problems 
depending  upon  principles  previously  determined  and 
classified  are: 

(1)  To  extend  the  curb  of  a  road  or  path  in  the  same 
straight  line  on  the  other  side  of  a  building  that  stands  on 
the  line  of  the  road,  using  the  method  of  146,  Exercise  10, 
or  of  146,  Exercise  11. 

(2)  To  find  the  distance  from  an  accessible  point  to  a 
point  on  the  other  side  of  a  stream,  wall  or  street;  (a)  by 
the  method  of  146,  Exercise  8,  or  (b)  by  the  method  of 
115,  Exercise  8. 

In  method  (a)  the  required  distance  is  measured  on  the 
ground,  while  in  method  (b)  the  plot  is  drawn  to  scale  on 
paper  and  the  required  distance  is  measured  on  the  drawing. 

(3)  The  plot  of  a  piece  of  ground  bounded  by  straight 


TRIANGLES 


89 


lines  may  be  made  by  the  method  of  146,  Exercise  6. 
Trees,  etc.,  may  be  located  by  any  of  the  methods  of  146, 
Exercise  5. 

(4)  To  lay  out  a  tennis  court,  basket-ball  court,  etc. 

(5)  Measure  the  angles  of  a  triangular  field,  in  order  to 
check  the  principle  of  94  (d)  and  119. 

FIELD  WORK 
153.  DEPARTMENT  OF  MATHEMATICS  -       —SCHOOL 

To  Lay  out  a  Tennis  Court 

Geometrical  Principles  Employed. — (1)  If  one  of  the 
equal  sides  of  an  isosceles  triangle  is  extended  through  the 
vertex  its  own  length  and  the  end  of  the  extended  line  is 
joined  to  the  nearer  end  of  the  base,  this  line  is  perpendicu- 
lar to  the  base  (104,  part  (6);  143,  Theorem  (6)). 
(2.)  Elementary  properties  of  a  rectangle. 


21 

19 

< 

3 

--^     / 

3                                            I 

4 

//^""" 

17 

^-^ 

IQ 

^-^^^           *"*•*• 

, 

^^^ 

^'""       A 

r                               / 

6 

/2"  -  ^ 

10 

t 

> 

22 

METHOD. — (1)    Set  corner  1;  set  stake  2  so  that  1-2  is 
parallel  to  boundary  line;  measure  1-3  =  36  feet. 

(2)  Erect  _L's  3-5  and  1-7;  measure  3-6  =  1-8=  78 
feet. 

(3)  Check  by  measuring  8-6  =  36  feet. 

(4)  Set  stakes  9,   10,   11,   12,  so  that  1-9=  10 -3  = 
etc.  =  4'  6". 

(5)  Set  stakes  13,   14,   15,   16,  so  that  9-13  =  18'  and 
9-14  =  60',  etc. 

(6)  Set  other  stakes  in  order. 


CHAPTER  VI 
POLYGONS 

PKINCIPLES  DETERMINED  EXPERIMENTALLY 

154.  Polygons    Defined. — A   polygon  is   a   plane   figure 
bounded  by  straight  lines. 

Almost  all  general  descriptive  words  are  used  in  the  same 
sense  for  polygons  as  for  triangles,  except  that  in  a  polygon 
having  an  even  number  of  sides,  two  angles  or  two  sides  are 
opposite,  while  in  a  triangle  and  in  polygons  having  an  odd 
number  of  sides,  a  side  and  an  angle  are  opposite. 

A  diagonal  is  a  straight  line  joining  two  non-consecutive 
vertices. 

A  convex  polygon  is  one  in  which  no  interior  angle  exceeds 
a  straight  angle. 

A  concave  polygon  is  one  in  which  at  least  one  interior 
angle  is  reflex. 

A  regular  polygon  is  both  equilateral  and  equiangular. 
Their  properties  are  considered  in  Chapters  XIV  and  XV. 

155.  Polygons  Named  According    to    the    Nximber    of 
Sides. 

A  quadrilateral  has  four  sides. 
A  pentagon  has  five  sides. 
A  hexagon  has  six  sides. 
A  heptagon  has  seven  sides. 
An  octagon  has  eight  sides. 
A  decagon  has  ten  sides. 
A  dodecagon  has  twelve  sides. 
A  pentadecagon  has  fifteen  sides. 

Other  polygons  may  also  be  named  from  the  Greek 
numeral  signifying  the  number  of  sides. 

90 


POLYGONS 


91 


156.  Classification  of  Quadrilaterals. 


quadrilateral 

f 


parallelogram 


rectangle 

square 
oblong 

oblique 
parallelogram 

rhombus 
rhomboid 

trapezoid 
trapezium 

Quadrilaterals  are  classified  according  to  the  parallelism 
of  opposite  sides: 

A  parallelogram  has  both  pairs  of  opposite  sides 

parallel. 

A  trapezoid  has  one  pair  of  opposite  sides  parallel. 
A  trapezium  has  no  sides  parallel. 
Parallelograms  are  classified  according  to  the  angles: 
A  rectangle  has  all  right  angles. 
An  oblique  parallelogram  has  no  right  angles. 
Rectangles  are  classified  according  to  relative  lengths  of 
sides : 

A  square  is  equilateral. 

An  oblong  has  one  pair  of  opposite  sides  longer  than 

the  other  pair. 

Oblique  parallelograms  are  classified  according  to  relative 
lengths  of  sides: 

A  rhombus  is  equilateral. 

A  rhomboid  has  one  pair  of  opposite  sides  longer 

than  the  other  pair. 

157.  Names  of  Special  Parts  of  Quadrilaterals. 
In  a  trapezoid : 

The  bases  are  the  parallel  sides. 
The  median  connects  the  mid-points  of  the  non- 
parallel  sides. 
The    altitude   is    a    perpendicular    to    the    bases, 

included  (comprehended)  between  them. 
An  isosceles  trapezoid  is  one  whose  non-parallel  sides  are 
equal. 

In  a  parallelogram : 

The  bases  are  the  horizontal  sides,  if  the  figure  is  so 


92  PLANE    GEOMETRY 

drawn;  but  either  pair  of  parallel  sides  may  be 
called  bases. 

The  altitude  is  a  perpendicular  to  the  bases,  com- 
prehended between  them. 

168.    EXERCISES 

1.  Draw  a  trapezium;  place  capital  letters  at  the  vertices.     Name 
the   quadrilateral,    reading   the   vertex  letters   consecutively. 
Draw  all  possible  exterior  angles;  all  possible  diagonals.     What 
parts  are  opposite? 

2.  Draw  a  pentagon,  ABODE.     What  angle  is  opposite  side  ABf 
What  side  is  opposite   /.Ef     What  angles  are  adjacent  to  BC? 
What  angle  is  included  by  BC  and  CD? 

3.  Draw  a  polygon  of  the  general  shape 
of  A  BCD.    Draw  a  polygon  congruent 
to  it;  a  polygon  similar  to  it.      Name 
one  set  of  homologous  sides  of  the 
three  polygons;  one  set  of  homologous 
angles.     Draw  homologous  diagonals 
in  each  polygon. 

4.  Draw  convex  polygons  of  3  sides ;  4  sides ;  5  sides ;  6  sides .     Draw 
(if  possible)  concave  polygons  of  3;  4;  5;  6  sides.     Draw  all 
possible  diagonals  of  these  polygons.     What  special  property 
has  at  least  one  diagonal  of  a  concave  polygon? 

5.  Draw  (approximately)  regular  polygons  of  3;  4;  5;  6  sides. 
Draw  (if  possible)  equilateral  polygons  which  are  not  equi- 
angular, and  also  equiangular  polygons  which  are  not  equi- 
lateral, of  3;  4;  5;  6  sides. 

6.  What  is  the  least  number  of  sides  a  polygon  may  have?  the 
greatest  number? 

7.  Draw  the  three  kinds  of  quadrilaterals.     Draw  the  four  kinds 
of  parallelograms. 

8.  What  other  names  are  given  to  a  regular  triangle?     What 
other  name  has  a  regular  quadrilateral? 

9.  Draw  a  trapezoid;  the  median;  the  altitude.     Name  the  bases; 
the  non-parallel  sides. 

10.  Draw  an  isosceles  trapezoid. 

11.  Draw  a  trapezoid  one  of  whose  non-parallel  sides  is  the  altitude. 

12.  Draw  a  parallelogram  whose  sides  form  its  altitudes. 

13.  Draw  the  two  altitudes  of  a  rhomboid. 

159.  Experiment  I. — The  properties  of  a  rhomboid. 
Draw  a  rhomboid  by  drawing  two  parallel  lines  cut  by 


POLYGONS  93 

two   other   parallel  lines  at   oblique  angles  and  unequal 
intercepts.     Draw  the  two  diagonals.     Observe: 

(a)  The  relation  between  opposite  sides. 

(6)  The  relation  between  opposite  angles. 

(c)  The  relation  between  consecutive  angles. 

(d)  The  various  pairs  of  congruent  triangles  formed. 

(e)  How  the  diagonals  divide  each  other. 
(/)  The  relative  length  of  the  diagonals. 
State  results. 

160.  Experiment  II. — The  properties  of  an  oblong. 
Observe,  in  addition  to  properties  (a),  (6),  (c),  (d),  (e)  of 

Experiment  I: 

(a)  That  there  are  four  pairs  of  overlapping  congruent 

triangles. 

(6)  The  relative  length  of  the  diagonals. 
State  results. 

161.  Experiment  III. — The  properties  of  a  rhombus. 
Observe : 

(a)  The  position  of  the  diagonals  with  respect  to  each 

other. 
(6)  How  the  diagonals  divide  the  angles  of  the  rhombus. 

162.  Experiment  IV. — The  properties  of  a  square. 
Observe  the  properties  of  the  rhomboid,  oblong,   and 

rhombus  which  belong  to  a  square. 

163.  Experiment  V. — Methods  of  constructing  a  rhom- 
boid. 

Construct  a  rhomboid : 

(a)  By  drawing  two  pairs  of  parallel  sides,  as  in  drawing 

the  figure  of  Experiment  I. 

(b)  By  drawing  two  pairs  of  equal  sides.     Complete  the 

figure  as  the  other  two  sides  fall. 


(c)  By  drawing  one  pair  of  sides  parallel   and  equal, 
and  completing  the  figure  as  the  other  two  sides  fall. 


94  PLANE    GEOMETRY 

(d)  By  drawing  the  two  diagonals,  oblique,  unequal,  and 
bisecting  each  other,  and  completing  the  figure  as 
the  four  sides  fall. 

164.  Experiment  VI.— Methods  of 
constructing. 

From  the  diagonals: 
(a)  An  oblong. 
(6)  A  rhombus. 
(c)  A  square. 

165.  Experiment  VII. — Properties  of  a  trapezoid. 
Observe  in  any  trapezoid : 

(a)  The  position  of  the  median. 

(6)  The  length  of  the  median  compared  with  the  lengths 
of  the  bases. 

(c)  How  the  median  divides  the  diagonals. 
Observe,  in  an  isosceles  trapezoid : 

(d)  Lengths  of  diagonals. 

(e)  Angle  properties  of  the  figure. 
(/)  Pairs  of  congruent  triangles. 


166.  EXERCISES 

1.  Do  the  diagonals  of  a  rhomboid  or  oblong  bisect  the  angles  of 
the  figure? 

2.  Are  the  diagonals  of  all  trapezoids  equal? 

3.  Do  the  diagonals  of  any  trapezoid  bisect  each  other? 

4.  Two  sides  of  a  rhomboid  are  21  and  36  feet  respectively;  what 
are  the  other  sides?     One  of  the  angles  is  70°;  what  are  the  other 
angles? 

5.  In  what  kind  of  quadrilaterals  are  the  diagonals  equal;  equal 
and  perpendicular;  unequal  and  perpendicular;  equal  without 
bisecting  the  angles  of  the  quadrilateral;  perpendicular  without 
bisecting  the  angles  of  the  quadrilateral;  axes  of  symmetry  of 
the  quadrilateral?     In  what  kind  may  only  one  diagonal  bisect 
the  angles  of  the  quadrilateral?     What  quadrilaterals  have  a 
center  and  axes  of  symmetry?     A  center  of  symmetry  but  no 
axes  of  symmetry? 

6.  The  bases  of  a  trapezoid  are  7  and  15  inches;  calculate  the 
median. 


POLYGONS  95 

167.  CONSTRUCTION  OF  POLYGONS  FROM  GIVEN  PARTS. 
EXERCISES 

Construct  the  following  figures: 

1.  A  pentagon   ABODE;  AB  =  1.5",     Z£  =  130°,    BC  =  1.0", 

ZC  =  90°,  CD  =  2.0",    ZD  =  150°,  DE  =  1.6". 

2.  A   pentagon   ABODE;   AB  =  1.2",    BC  =  2.8",    CD  =  0.7", 
DE  =  3.0",  EA  =  2.2";  diagonals,  5#  =  2.7",  CE  =  3.4". 

3.  A  hexagon  in  which  each  side   =  1.0",  and  each  angle  ==  120°. 

4.  A  parallelogram  of  which  two  sides  and  the  included  angle  are 
to  be  equal  respectively  to  two  given  sects  and  to  one  given 
angle. 

5.  A  parallelogram  of  which  two  adjacent  sides  and  a  diagonal 
are  given. 

6.  A  parallelogram  of  which  one  side,  one  diagonal,  and  the  angle 
included  between  the  given  side  and  diagonal  are  given. 

7.  A  parallelogram  of  which  two  diagonals  are  respectively  3.0" 
and  1.7",  and  their  included  angle  is  50°. 

8.  A  parallelogram  of  which  one  side,  one  diagonal,  and  the  angle 
included  between  the  two  diagonals  are  given. 

9.  A  parallelogram  of  which  one  side  =  1.6",  a  diagonal  =4.0",  the 
angle  included  by  the  two  diagnals  =  45°. 

10.  A  rectangle  of  which  two  adjacent  sides  are  given. 

11.  A  rectangle  of  which  one  side  and  one  diagonal  are  given. 

12.  A  rectangle  of  which  one  diagonal  and  the  angle  between  the 
diagonals  are  given. 

13.  A  rectangle  of  which  one  diagonal  and  an  angle  between  the 
diagonal  and  a  side  are  given. 

14.  A  square  of  which  one  side  =  2.5". 

15.  A  square  of  which  a  diagonal  =  1.7". 

16.  A  rhombus  of  which  the  two  diagonals  are  respectively  3.0" 
and  2.4". 

17.  A  rhombus  of  which  one  side  and  one  diagonal  are  given. 

18.  A  rhombus  of  which  one  side  =  2.0",  and  one  diagonal  =  4.5". 

19.  A  trapezoid  of  which  the  two  bases  and  the  altitude  are  given. 

20.  A  trapezoid  of  which  one  base,  the  two  adjacent  angles,  and 
altitude  are  given. 

21.  An  isosceles  trapezoid  of  which  one  base,  one  of  the  non-parallel 
sides,  and  an  angle  adjacent  to  the  given  base  are  given. 

22.  A  trapezoid  of  which  the  median,  lower  base,  and  altitude  are 
given. 

23.  A  trapezium  of  which  the  four  sides  and  their  order,  and  a 
diagonal  (and  its  position  in  the  figure)  are  given 


96  PLANE    GEOMETRY 

24.  A  trapezium  of  which  the  two  diagonals  and  their  included 
angle  are  given. 

25.  A  trapezium  of  which  the  four  sides  and  their  order  and  one 
angle  are  given. 

26.  A  trapezium  of  which  the  four  sides  in  order  are  2.0",  1.5", 
2.4",  3.0",  and  the  angle  included  by  the  2.0"  and  3.0"  sides 
is  80°. 

27.  Which  of   the   preceding  constructions   are   ambiguous    (two 
possible  figures);   which  indefinite    (any  number  of   possible 
figures);  which  impossible? 


CHAPTER  VII 
POLYGONS 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

168.  The  Second  Step  in  Classification. — No  additional 
postulates  are  required  in  the  analyses  of  the  principles  of 
Chapter  VI,  which  depend  upon  the  principles  of  parallels 
(Chapter  III)  and  of  congruent  and  isosceles  triangles  (Chap- 
ter V) .     The  polygons  of  most  importance  (other  than  tri- 
angles) are  parallelograms  and  trapezoids.     In  the  Theorems 
which  follow,  the  analysis  of  which  depends  upon  congruence 
of  triangles,  draw  the  triangles  separately  and  overlay  the 
equal  parts  with  colored  crayons. 

169.  Theorem  I.— A  diagonal  of  a  parallelogram  divides 
it  into  two  congruent  triangles. 

Helps. — Draw  a  rhomboid.  Show  that  three  parts  of  the 
triangles  are  respectively  equal  (116). 

170.  Theorem   II. — (a)    Any   two   opposite   angles   of  a 
parallelogram  are  equal;  (b)  any  two  consecutive  angles  are 
supplementary . 

Helps.— (a)  This  follows  from  169,  or  from  81.  (b) 
This  follows  from  79,  Theorem  (l). 

171.  Theorem  III. — The  opposite  sides  of  a  parallelogram 
are  equal.  c  D 

Corollary. — Parallels  are  everywhere 
equally  distant. 

Helps.— (1)  AC\\BD,  why?     (2)  AC    _ 
=  BD,  why?  A  B 

172.  Theorem   IV. — The   diagonals   of   a   parallelogram 
bisect  each  other. 

Help. — Be  careful  to  select  the  triangles  whose  congru- 
ence will  prove  that  the  parts  of  the  diagonals  are  respec- 
tively equal. 

7  97 


98  PLANE    GEOMETRY 

173.  Theorem  V. — The  diagonals  of  a  rectangle  are  equal. 
Help. — The  triangles  selected  for  congruence  must  have 

the  entire  diagonals  for  their  respective  sides. 

174.  Theorem  VI. — The  diagonals  of  a  rhombus  (a)  are 
perpendicular,  and  (b)  bisect  the  angles  of  the  figure. 

Help.— Select  two  of  the  small  adjacent  triangles  for 
congruence;  or  apply  properties  of  an  isosceles  triangle. 

175.  Theorem  VII. — State  and  analyze  all  the  properties 
of  a  square. 

CONVERSE  PRINCIPLES 

176.  Theorem  VIII. — //  both  pairs  of  opposite  sides  of  a 
quadrilateral  are  equal,  the  figure  is  a  parallelogram  (163, 
part  (6)).  A  ,B 


c/ 

Hypothesis. — Quadrilateral  ABDC  with  AB  =  CD  and 
AC  =  BD. 

Conclusion. — ABDC  is  a  parallelogram;  that  is,  AB\\CD 
and  AC\\BD. 

Analysis. 

STATEMENT  REASON 

1.  Draw  diagonal  BC  Postulate 

2.  Compare  &ABC  and  BDC 
3. 


4. 


Find     three     respectively 


_       equal  parts 
o. 


Give  reasons 


6.  .'.  A  are  ^  What  postulate? 

7.  .'.  Zl  =  Z4  and  Z3  =  Z2         Horn,  parts  of  ^  A 

8.  .'.  AC\\BD  and  AB\\CD  72 

177.  Theorem  IX. — //  one  pair  of  opposite  sides  of  a 
quadrilateral  are  parallel  and  equal,  the  figure  is  a  paral- 
lelogram. 

Helps. — Construct  the  figure  as  in  163,  part  (c).  The 
figure  may  be  proved  to  be  a  parallelogram  by  showing  that 


POLYGONS  99 

both  pairs  of  sides  are  parallel,  each  to  each,  as  in  176;  or 
by  showing  that  both  pairs  of  sides  are  equal,  each  to  each, 
since  we  may  apply  the  principle  of  176. 

178.  Theorem   X. — //  the   diagonals   of  a   quadrilateral 
bisect  each  other,  the  figure  is  a  parallelogram. 

Helps. — Both  pairs  of  opposite  sides  may  be  proved 
parallel,  or  the  principles  of  either  176  or  177  may  be 
applied.  Of  these,  the  last  mentioned  makes  the  best 
analysis. 

179.  Theorem  XI. — //  the  diagonals  of  a  quadrilateral 
bisect  each  other  and  are  equal,  the  figure  is  a  rectangle. 

Helps. — (1)  The  figure  is  a  parallelogram  (178);  (2) 
prove  two  overlapping  triangles  are  congruent;  (3)  two 
consecutive  angles  of  the  figure  are  equal;  (4)  these  angles 
are  also  supplementary;  (5)  and  are  therefore  both  right 
angles. 

180.  Theorem  XII. — A  line  which  bisects  two  sides  of  a 
triangle  (a)  is  parallel  to  the  third  side;  and  (b)  equals  one-half 
the  third  side. 


Hypothesis. — 
Conclusion . — (a) . 
(6). 
Analysis. 

STATEMENT  REASON 

1.  Extend  DE]  draw  CF\\BA 

2.  AADE  Q*  AEFC 

3.  DB\\  and  =  CF 

4.  DFCB  is  a  parallelogram 

5.  .*.  DE\\BC 

6.  DE  =  EF 

7.  .'.  DE  =      BC 


100  PLANE    GEOMETRY 

181.  Theorem  XIII. — A  line  which  bisects  one  side  of  a 
triangle  and  is  parallel  to  a  second  side,  bisects  the  third  side. 

A 


Helps.— (I)  Extend  DE  and  draw  CF\\BA;~ (2)  DFCB  is 
a  parallelogram;— (3)  BD  =  CF,—(4)  AADE  *  AEFC: 
—(5)  /.  AE=  EC. 

182.  Theorem   XIV. — A    line   which   bisects   one   of  the 
non-parallel  sides  of  a  trapezoid  and  is  parallel  to  the  bases 
bisects  the  other  non-parallel  side. 
A  B 


7 


H 

Helps.— (1)  Draw£G  smdFH\\AD;—(2)  ABGF  *  AFHC; 
—(3)  .%  BF  =  FC. 

183.  Theorem  XV. — A  line  which  joins  the  middle  points 
of  the  two  non-parallel  sides  of  a  trapezoid  (i.e.,  the  median) 
is  parallel  to  the  bases. 

A  B 


Helps.— (1)  Draw  EG\\AB  and  DC;—(2)G  is  middle  point 
of  BC;— (3)  F  and  G  coincide  (63,  Postulate  5);— (4)  .'.  EF 
and  EG  coincide  (63,  Postulate  1);— (5)  /.  EF\\AB  and  DC 
(127,  Postulate  21). 

184.  Theorem  XVI.— The  median  of  a  trapezoid  equals 
one-half  the  sum  of  the  bases. 

Helps.— (1)  Draw  a  diagonal;  (2)  the  part  of  the  median 
intercepted  within  each  triangle  equals  J  a  base;  etc. 


POLYGONS 


185.  Theorem  XVII.  —  The  medians  of  a  triangle  are 
concurrent  in  a  point  which  is  two-thirds  the  length  of  each 
median  from  its  vertex. 


Helps. — Part  I. — (1)  BE  and  CF  intersect  in  point  0; 
—(2)  Draw  EF',  draw  GH  bisecting  BO  and  CO  respectively; 
-(3)  EF  is  ||  and  =  to  GH;—  (4)  draw  FG  and  EH;—  (5) 
FEHG  is  a  parallelogram;— (6)  .'.'  OF  =  OH  and  OE  =  OG; 
-(7)  /.  BO  =  f  BE  and  CO  =  f  CF. 

Part  II.  (1)  Similarly  BO'  =  f  BE  and  AO'  =  f  AD; 
—(2)  .'.  points  0  and  0'  coincide,  and  the  medians  are 
concurrent. 

Notice  that  this  is  the  first  of  the  concurrent  lines  of  a 
triangle  (111,  part  (&))  which  has  been  analyzed. 

186.  The  Third  Step-  in  Classification. — The  following 
properties  of  polygons  (187  and  188)  might  never  have  been 
discovered  if  we  were  forced  to  rely  upon  experimental 
methods  of  investigation  alone  (see  142). 

187.  Theorem  XVIII. — The  sum  of  the  interior  angles  of  a 
polygon  depends  upon  the  number  of  sides  of  the  polygon  and 
is  independent  of  its  shape. 

Draw  polygons  of  different  numbers  of  sides  and  of  differ- 
ent shapes;  let  the  number  of  sides  of  each  polygon  be 
denoted  by  n. 

Helps. — (1)  Mark  any  point  P,  within  each  polygon, 
and  draw  auxiliary  lines  from  P  to  each  vertex;  (2)  state 
the  number  of  triangles  formed  in  each  polygon  in  terms 
of  n;  (3)  state  the  sum  of  the  angles  of  one  triangle  using 
the  straight  angle  as  the  unit;  (4)  state  the  sum  of  the 
angles  of  all  the  triangles  in  terms  of  n;  (5)  deduct  the  sum 
of  the  angles  around  point  P;  (6)  the  resulting  expression 
gives  the  sum  of  the  interior  angles  of  the  polygon  in 


102 


PLANE    GEOMETRY 


terms  of  n,  the  number  of  sides.  Express  the  result  also 
in  degrees. 

188.  Theorem  XIX. — The  sum  of  the  exterior  angles  of  a 
polygon,  one  at  each  vertex,  is  a  constant  value  and  is  inde- 
pendent of  the  number  of  sides  of  the  polygon. 

Helps. — (1)  State  the  value  of  the  sum  of  an  interior 
angle  and  an  exterior  angle  at  each  vertex  of  the  polygon, 
using  the  straight  angle  as  the  unit ;  (2)  state  the  sum  of  all 
the  interior  and  of  all  the  exterior  angles  in  terms  of  n; 

(3)  state  the  sum  of  all  the  interior  angles  alone  (187); 

(4)  subtract;  the  difference  is  the  value  of  all  the  exterior 
angles.     Express  the  result  also  in  degrees. 

189. — THEOREM. — Every  triangle  is  isosceles. 
This  theorem  is  found  in  Mathematical  Recreations  and 
Problems,  by  W.  W.  R.  Ball ;  published  by  Macmillan  and  Co. 


cBS             L 

?  / 

\c 

\  * 

\\ 
x  \ 
\^x 

v 

/ 
; 
/ 

/ 
/          t 

V 

0 

/>£ 

//  x 

/  / 

'/ 

Hypothesis. — Any  A  ABC. 
Conclusion. — A  ABC  is  isosceles. 
Analysis. 

STATEMENT 

1.  Draw  DG  the  J. -bisector  of  BC, 
AO  bisecting   ZA,  BO  and  CO, 
GEL  AC,  OF  _L  AB. 

2.  There  are  four  possible  cases: 
(a)  DG  and  AO  do  not  intersect, 


REASON 
Postulate 


POLYGONS  103 

(b)  DG  and  AO  intersect  at  point  Z>, 

(c)  DG  intersects  AO  within  AA5C, 

(d)  DG  intersects  AO  outside  AABC. 
Consider  Case  (a) : 

3.  Then  DG\\AO 

4.  .\AO±BC 

5.  .'.  A  ABC  is  isosceles  134.  Theorem  (1) 
Consider  Case  (6) : 

6.  Then  AO  bisects  BC 

7.  .*.  AABC  is  isosceles  134.  Theorem(2) 
Consider  Case  (c)  (Figure  1) : 

8.  Compare  &AOF  and  AOE 

9.  AO  =  AO 

10.  x  =  y 

11.  s  =  t 

12.  .'.  AAOF  £  A40#  Postulate  19 

13.  .'.  AF  =  AE  and  OF  =  OE 

14.  Compare  A50F  and  COE 

15.  OB  =  OC 

16.  OF  =  OE 

17.  u  =  w 

18.  .'.  A£OF  ^  C0#  Postulate  17 

19.  .'.  BF  -  C£7 

20.  /.  AF  +  BF  =  AE  +  C#  Axiom 

21.  Or  AB  =  AC 

22.  .'.  AABC  is  isosceles 
Consider  Case  (d)  (Figure  2) : 

23.  AAOF  ±  AAOE 

24.  .'.  AF  =  AE  and  OF  =  0£; 

25.  ABOF  ^  C0£/ 

26.  .'.  BF  =  CE 

27.  .'.  AF  -  BF  =  AE  -  CE 

28.  Or  AB  =  AC 

29.  .'.  AABC  is  isosceles 

Thus  in  every  case  A  ABC  is  isosceles.  It  has  therefore 
been  proved  that  no  triangle  other  than  an  isosceles  triangle 
can  be  drawn. 


104 


PLANE    GEOMETRY 


190.  The  Fourth  Step  in  Classification. — The  principles 
of  187,  188  and  189,  which  were  discovered  by  making  use 
of  methods  of  analysis,  and  not  by  experiment,  as  were  the 
principles  of  Chapters  II,  IV  and  VI,  must  now  be  tested 
experimentally,  and  verified  or  rejected  as  the  result  of  such 
tests  shall  indicate. 

EXERCISES 

1.  Draw  polygons  of  4  sides,  5  sides,  6  sides,  7  sides  and  8  sides, 
respectively,  of  different  shapes,  both  convex  and  concave. 
Draw  one  exterior  angle  at  each  vertex.  Note  that  the  exterior 
angle  is  negative  at  a  vertex  where  the  interior  angle  is  reflex. 
Measure  with  the  protractor  each  interior  and  each  exterior 
angle  of  all  the  polygons.  Find  the  sum  of  the  interior  and  of 
the  exterior  angles  for  each  figure,  and  enter  the  results  in  such 
a  table  as  the  following: 


Number  of  sides 
of  polygon 

Sum  of  interior 
angles  measured 
in  straight  angles 

Sum  of  exterior 
angles  measured 
in  straight  angles 

4 

5 

6 

7 

8 

Does  the  result  which  may  now  be  stated  for  the  general 
polygon  of  n  sides,  agree  with  the  values  deduced  in  187  and  188? 
2.  May  triangles  other  than  isosceles  triangles  be  drawn?  Does 
the  conclusion  of  the  theorem  of  189  agree  with  experiment? 
Do  you  find  any  flaw  in  the  logical  analysis?  Either  we  must 
discover  wherein  the  logical  error  is  concealed,  or  our  entire 
logical  system  may  be  invalidated. 

The  error  will  be  found  in  having  assumed  incorrect,  or  im- 
possible, positions  for  the  point  0.  Such  an  assumption  is  a  Pos- 
tulate; which  having  been  found  to  be  false  must  be  rejected  or 
replaced  by  a  correct  Postulate.  An  accurate  construction  of 
the  figure  will  reveal  the  nature  of  the  error.  State  as  a 
Postulate,  thus  discovered  experimentally,  the  principle:  If 
perpendiculars  are  drawn  from  the  point  of  intersection  of  an 
angle-bisector  and  the  perpendicular-bisector  of  the  opposite  side 


POLYGONS  105 

of  a  scalene  triangle,  to  the  other  two  sides  of  the  triangle,  these 
perpendiculars  fall,  etc. 

ADDITIONAL  THEOREMS 

191.  (1)  If  the  diagonals  of  a  quadrilateral  are  perpendicular- 
bisectors  of  each  other,  the  figure  is  an  equilateral  parallelogram. 

(2)  //  the  diagonals  of  a  quadrilateral  are  perpendicular- 
bisectors  of  each  other,  and  equal,  the  figure  is  a  square. 

(3)  //  two  straight  lines  are  equally  distant  from  each  other 
at  any  two  points   (measured  on  perpendiculars)   they  are 
parallel. 

(4)  Analyze  Theorem  XVII,  187,  by  dividing  the  polygons 
into  triangles  by  drawing  diagonals,  but  so  that  no  triangles 
overlap. 

(5)  Each  interior  angle  of  an  equiangular  polygon  can  be 
expressed  in  terms  of  the  number  of  sides  of  the  polygon. 

(6)  Each  exterior  angle  of  an  equiangular  polygon  can  be 
expressed  in  terms  of  the  number  of  sides  of  the  polygon. 

(7)  The  lines  joining  in  order  the  middle  points  of  the  sides 
of  any  quadrilateral  form  a  parallelogram. 

Helps. — Draw  one  or  both  diagonals  of  the  given  quadri- 
lateral. Apply  180  (a)  or  (6).  Note  the  three  possible 
variations  in  the  analysis. 

(8)  The  lines  joining  the  middle  points  of  the  opposite  sides 
of  any  quadrilateral  bisect  each  other. 

(9)  The  lines  joining  the  middle  points  of  the  opposite  sides 
of  an  oblong  form  a  rhombus. 

Help. — The  new  figure  must  be  shown  to  be  equilateral, 
and  its  diagonals  to  be  unequal. 

(10)  The  diagonals  of  an  oblique  parallelogram  are  unequal. 
Help.— Apply  143,  Theorem  (33). 

(11)  A    parallelogram   whose   diagonals   are    unequal   is 
oblique.  A  E          B 

(12)  In  the  parallelogram  ABCD,  E 
and  F  are  the    mid-points  of  AB  and 
DC.j^Then  DE  is  parallel  and  equal 
toFB. 


106 


PLANE    GEOMETRY 


(13)  In  the  figure  of  Theorem  (12),  the  lines  DE  and  FB 
trisect  the  diagonal  AC.  A         c         R 

Helps.— (1)  In  &ABH,  GE  bisects 
AH;  (2)  similarly  FH  bisects  GC;  (3)         / /G 
/.  AG  =  GH  =  HC.  //     > 

(14)  The  bisectors  of  the  angles  of  a    D 
quadrilateral  form  another  quadrilateral  whose  opposite  angles 
are  supplementary. 

(15)  Corollary  of  Theorem  (14).     When  the  given  quadri- 
lateral is  a  parallelogram,  that  formed  by  the  angle  bisectors 
is  a  rectangle. 

(16)  A  convex  polygon  cannot  have  more  than  three  acute 
angles. 

Help.—li  an  interior  angle  is  acute,  what  is  the  exterior 
angle  at  that  vertex? 

(17)  In  a  triangle  ABC,  AD  is  a  median  prolonged  to  E 
so  that  DE  =  AD.     Then  CE  is  parallel  to  AB. 

(18)  The  angles  adjacent  to  either  base  of 
an  isosceles  trapezoid  are  equal. 

(19)  The  diagonals  of  an  isosceles  trape- 
coid  are  equal. 

(20)  The  sum  of  the  angles  at  the  points 
of  a  five-pointed  star  is  a  straight  angle. 

(21)  The    perpendiculars    drawn  from 
opposite  vertices  of  a  parallelogram  to  a 
diagonal  are  equal. 

(22)  Any  line  drawn  through  the  point  of  intersection  of  the 
diagonals  of  a  parallelogram,  and  terminated  by  the  sides  is 
bisected. 

(23)  //  a  diagonal  of  a  parallelogram  bisects  an  angle,  the 
parallelogram  is  equilateral. 

(24)  Analyze  Theorem 
XVII,     185,     as    follows: 
Draw  AABC  and  medians 
BE  and  CF  intersecting  at 
0;   draw  AD   through    0. 
Prove    that   AD   is  a  me- 


POLYGONS 


107 


dian;  i.e.,  BD  =  BC;  and  that  0  is  a  trisection  point  on 
each  median. 

Helps.— (1)  Prolong  AD  so  that  OG  =  AO  and  draw  GB 
and  GC;— (2)  in  AAGC,  OE\\GC;—(3)  in  AABG,  OF\\BG; 
-(4)  .'.  OCGB  is  a  parallelogram;— (5)  in  AAGC,  OE  = 
J  GC,— (6)  .'.  OE  =  |  BO  =  £  BE;— (7)  BD  =  DC. 

(25)  //  £/ie  angles  adjacent  to  either  base  of  a  trapezoid  are 
equal,  the  trapezoid  is  isosceles.  ^ 

(26)  //  the  diagonals  of  a  trapezoid     

are  equal,  the  trapezoid  is  isosceles.  . 

Help. — Draw   the   altitudes  from      — \L 
points  A  and  B. 

(27)  Any  obtuse  angle  equals  a  right  angle. 
Hypothesis. — A  right  /.CDA   and 

an  obtuse  /.FAD. 

Conclusion.—  /.CDA  =    /.FAD. 

Helps. — (1)  Construct  rectangle 
ABCD,  extend  CB  to  F,  lay  off  AE 
=  AB,  draw  CE,  draw  HO  perpen- 
dicular-bisector of  CB  and  KO  per- 
pendicular-bisector of  CE,  draw  OD, 
OC,  OA,  OE. 

(2)  HO  intersects  KO  at  0;  143,  Theorem  (3.) 

(3)  ACDO  *  AEAO;  3  sides   respectively  equal;    116, 
Postulate  15. 

(4)  .'.   ZCDO  =    ^EAO. 

(5)  ZADO  =    ZDAO;  base  Zs  of  isosceles  A;  121. 

(6)  /.   ZCDA  =    /.FAD;  64,  Axiom  3. 

Find  the  error  in  the  preceding  theorem,  which  may  be 
found  in  Mathematical  Recreations  and  Problems  by  W.  W. 
R.  Ball;  published  by  Macmillan  and  Co. 

192.  REVIEW  EXERCISES 

1.  Why  was  it  impossible  to  include  the  analyses  of  180,  181  and 
186  in  the  chapter  on  triangles  (Chapter  V)  ? 

2.  Two  consecutive  sides  of  a  parallelogram  are  10  and  12  inches 
respectively,   and  the  included  angle  is   75°.     Calculate  the 
remaining  sides  and  angles. 


108  PLANE    GEOMETRY 

3.  The  bases  of  a  trapezoid  are  7  inches  and  10  inches.     Calculate 
the  length  of  the  median. 

4.  The  median  of  a  trapezoid  is  13.5"  and  one  base  is  7.4".     What 
is  the  other  base?     Write  a  formula  for  the  median  m,  in  terms 
of  the  bases  &i  and  62. 

5.  Calculate  the  sum  of  the  interior  angles  of  a  triangle,  quadri- 
lateral, pentagon,  hexagon,  octagon,  decagon,  using  the  formula 
of  187. 

6.  Calculate  the  value  of  each  interior  angle  and  of  each  exterior 
angle  of  equiangular  polygons  of  3,  4,  5,  6,  8  and  10  sides. 

7.  If  one  of  the  interior  angles  of  an  equiangular  polygon  is  150°, 
how  many  sides  has  the  polygon? 

8.  Calculate  the  number  of  sides  of  an  equiangular  polygon  if 
one  of  the  exterior  angles  is  18°. 

9.  How  many  sides  has  a  polygon  if  the  sum  of  the  interior  angles 
is  ten  straight  angles?     If  the  sum  of  the  interior  angles  is  630°? 

10.  How  many  sides  has  a  polygon  the  sum  of  whose  interior  angles 
equals  the  sum  of  the  exterior  angles? 

11.  How  many  sides  has  a  polygon  the  sum  of  whose  interior  angles 
is  three  times  the  sum  of  the  exterior  angles?     a  times  the  sum 
of  the  exterior  angles? 

12.  Five  of  the  interior  angles  of  a  hexagon  are  70°,  135°,  140°,  130°, 
45°  respectively.     What  is  the  other  angle? 

•  13.  Under  what  circumstances  should  a  postulate,  which  is  sup- 
posed to  state  a  geometric  principle  of  the  most  elementary 
and  most  evident  kind,  be  modified  or  rejected? 

14.  Can  an  argument  based  logically  upon  the  accepted  postulates 
and  theorems  of  a  science  prove  a  principle  which  is  found  to  be 
contrary  to  the  fact  as  established  by  experimental  investiga- 
tion?    Which  possesses  the  higher  degree  of  certainty,  logically 
deduced  or  experimental  laws? 

15.  Find  out,  if  possible,  what  was  generally  accepted  by  the  scien- 
tists of  the  world  as  the  law  of  falling  bodies  before  Galileo  made 
his  famous  experiments  at  the  Tower  of  Pisa  about  1590.     Was 
the  logically  deduced  "law"  confirmed  or  destroyed  as  the  result 
of  these  experiments? 

16.  Construct  a  regular  hexagon  each  side  of  which  is  1  inch.     Cal- 
culate the  angles,  and  draw  a  side  and  measure  an  angle  alter- 
nately. 

17.  In  the  story  of  Sir  Galahad  will  be  found  a  description  of  the 
"pen tangle"  which  adorned  his  shield.     Design  the   "pent- 
angle"  from  the  description. 

18.  Make  a  list  of  all  theorems  of  Chapters  III,  V  and  VII  in  which 
the  equality  of  two  or  more  sects  has  been  shown  to  be  a  neces- 


POLYGONS  109 

sary  consequence  of  the  constructed  property  of  the  figure. 
Make  a  similar  list  concerning  equal  angles. 

19.  State  theorems  of  congruent  parallelograms  and  quadrilaterals, 
in  terms  of  the  required  number  and  arrangement  of  respec- 
tively equal  sides  and  angles. 

20.  Show  experimentally  that  the  relation  between  the  number  of 
sides  of  a  polygon  and  the  total  number  of  possible  diagonals 
is  expressed  by  the  formula,   d  =  (n  —  l}n  —  [3  +  (3  +  4  + 
.    .    .   n)];  where  d  =  number  of  diagonals  and  n  =  number  of 
sides. 


£. 


B 


21.  Why  is  the  edge  AB  of  the  parallel  ruler  shown  in  the  figure, 
parallel  to  the  edge  CD  in  all  positions? 

APPLICATIONS 

193.  Physics  and  Engineering.  The  Parallelogram  of 
Forces. — When  several  forces  act  simultaneously  upon  the 
same  body  they  may  be  replaced  by  a  single  force  which  will 
produce  an  effect  equivalent  to  the  combined  effect  of  the 
acting  forces.  This  single  equivalent  force  is  the  resultant; 
the  acting  forces  are  the  components.  The  law  of  the  rela- 
tion of  resultant  and  two  component  forces  may  be  stated 
geometrically  as  follows:  If  the  sides  of  a  parallelogram 
are  drawn  in  units  representing  the  values  of  the  component 
forces,  the  angle  between  the  sides  being  the  angle  between 
the  directions  of  the  forces,  then  the  length  and  direction 
of  the  diagonal  of  the  parallelogram  represents  the  value 
and  direction  of  the  resultant.  This  law  is  determined 
experimentally. 


110  PLANE    GEOMETRY 

EXERCISES 

1.  What  lines   in   the   figure  represent   the   component   forces? 
What  line  represents  the  resultant? 

2.  Two  forces  of  10  and  20  Ib.  respectively,  act  at  an  angle  of  45° 
to  each  other.     Construct  the  parallelogram  of  forces  to  scale 
and  find  the  resultant  and  the  angle  which  the  resultant  makes 
with  the  direction  of  the  larger  force. 

3.  The  resultant  of  two  forces  is  30  Ib.,  and  makes  an  angle  of  30° 
with  one  of  its  components  which  has  a  value  of  40  Ib.     Find 
the  value  of  the  other  component  and  the  angle  between  the 
components. 

4.  Show  how  the  resultant  of  three  given  forces  may  be  obtained 
by  combining  the  resultant  of  two  of  them  with  the  third.     May 
this  combination  be  made  in  any  order? 

5.  Let  the  resultant,  r,  of  three  forces,  4 
and  two  of  these  forces,  a  and  6,  be       ^  / 
given  in  value  and  direction.     Find        / 

the  value  and  direction  of  the  third  D  L - w 

component. 

The    parallelogram    of  velocities    is 
constructed  in  the  same  way  as  for  forces. 

6.  A  boat  is  travelling  northward  at  a  velocity  of  20  miles  per 
hour,  and  a  passenger  is  walking  across  the  deck  towards  the 
east  at  a  velocity  of  4  miles  per  hour.     Find  the  actual  direction 
and  velocity  of  the  passenger  relative  to  a  fixed  object. 

7.  A  mail  bag  is  thrown  from  a  train  at  a  right  angle  to  the  track, 
at  a  velocity  of  10  feet  per  second.     The  train  is  travelling 
50  miles  per  hour.     Find  the  direction  and  velocity  with  which 
the  mail  bag  travels  until  it  hits  the  ground. 

Help. — The  velocity  of  the  train  is  equivalent  to  88  feet  per 
second. 

8.  An  aeroplane  is  travelling  80  miles  per  hour  in  a  northeasterly 
direction,  while  the  wind  is  blowing  from  the  east  at  a  velocity 
of  40  miles  per  hour.     Find  the  direction  and  velocity  in  relation 
to  the  earth  at  which  the  aeroplane  is  travelling. 

9.  A  vessel  while  moving  at  a  rate  of  30  miles  per  hour  parallel 
to  the  shore  line,  fires  a  shot  at  a  target  which  is  exactly  abeam. 
The  projectile  has  a  velocity  of  200  feet  per  second.     In  what 
direction  must  the  gun  be  aimed  if  the  shot  is  to  hit  the  target? 

The  Polygon  of  Forces. — Several  forces  acting  simultane- 
ously upon  the  same  body  may  balance  each  other  so  that 
the  resultant  effect  is  zero.  The  forces  are  then  said  to  be 


POLYGONS 


111 


in  equilibrium.  The  geometric  condition  for  equilibrium 
of  a  system  of  forces  is  that  the  lines  representing  in  length 
and  direction  the  values  and  directions  of  the  forces  shall 
form  the  sides  of  a  polygon  when  placed  end  to  end  in  the 
same  direction.  The  forces  a,  b,  c  in  Fig.  1  are  in  equili- 
brium if  they  form  a  triangle  as  in  Fig.  2. 


FIG.  1. 


FIG.  2. 


EXERCISES 

10.  It  is  evident  that  the  force  c  balances  the  resultant  r  of  forces 
a  and  b.     Show  by  construction  that  force  a  balances  the  result- 
ant of  forces  b  and  c,  and  also  that  force  b  balances  the  resultant 
of  forces  a  and  c. 

11.  Draw  a  point  P  and  any  four  forces  which  are  in  equilibrium 
acting  at  the  point  P,  no  two  forces  being  in  the  same  straight 
line. 

The  principle  of  the  polygon  of  forces  is  employed  by 
engineers  in  determining  the  breaking  stress  in  framed 
structures  such  as  bridge  and  roof 
trusses,  buildings  and  cranes,  (see 
figures  in  150).  From  the  stress  as  2, 

thus  found,  the  size  of  the  piece,  or 
member,  is  calculated. 


12.  Let  the  bridge  truss  shown  be  designed  to  support  a  total  uni- 
form load  of  40,000  Ib.  A  force  of  10,000  Ib.  may  then  be  con- 
sidered as  supporting  each  end.  The  remaining  20,000  Ib. 
rests  upon  the  abutments  and  has  no  effect  upon  the  truss 
stresses.  Draw"  the  polygon  (triangle)  of  forces  at  the  point 
A,  where  the  value  and  direction  offeree  x  and  also  the  direc- 


112  PLANE    GEOMETRY 

tions  of  forces  y  and  z  are  known.     Determine  the  amount  of 
stress  in  both  the  upper  and  lower  bridge  members. 

194.  Surveying. — Many  uses  are  found  for  the  principles 
of  polygons. 

EXERCISES 

1.  A  surveyor  has  measured  three  angles  of  a  four-sided  field: 
75°  30.5',  128°  51.2',  135°  3.7'.     If  these  are  correct,  what  is 
the  fourth  angle? 

2.  A  surveyor  has  measured  the  angles  of  a  five-sided  field;  110° 
21.7',  141°  13.3',  148°  54.0',  47°  20.9',  92°  10.6'.     What  total 
error  has  been  made?     Distribute  the  error  equally  among  the 
measured  angles  to  find  the  most  probable  correct  values. 

3.  If  the  four  sides  of  a  quadrilateral  field  are  measured  with  a  tape 
may  a  map  of  the  field  be  drawn  to  scale?     What  more  must  be 
measured?     If  no  transit  is  available  what  other  line  measure- 
ment will  determine  the  field? 

'4.  Explain  the  construction  of  a  line  through  point  C  parallel  to 
line  AB,  by  an  adaptation  of  the  principles  of  163,  part  (d),  and 
178.  In  an  out-of-doors  problem  line  AB  will  be  a  street  curb, 
fence,  etc.,  and  a  stake  is  to  be  driven  at  a  point  E  so  that  CE 
is  parallel  to  AB. 

C  E 


A  B 

5.  A  surveyor  wishes  to  determine  the  line  AF,  point  F  being  visible 
from  point  A.  Explain  the  method.  What  parallelogram  prop- 
erty is  made  use  of  in  finding  AF? 


C  D 

6.  Another  method  of  laying  out  a  line  through  point  C  parallel  to 
AB  is  shown  by  the  figure.  Explain  how  the  measurements  are 
made  and  upon  what  principle  it  depends. 

D 


B 


POLYGONS 


113 


195.  Designing. — Triangles,  parallelograms  and  regular 
polygons  appear  in  a  great  variety  of  artistic  designs  in 
which  symmetry  is  also  an  essential  feature. 

EXERCISE 

The  following  designs  are  for  tiling,  parquetry  flooring,  lino- 
leums and  embroidery.  Draw  some  of  these  designs  to  scale,  or 
copy  or  originate  other  similar  designs.  Suitable  coloring  may 
be  used. 


XXXX 


XXXX 


MMMX 


A    A    A 


UHt 


I-I  I 


196.  PROBLEMS  FOR  FIELD  WORK.— In  these  problems  the 
principles  of  Chapters  VI  and  VII  are  employed. 

1.  To  lay  out  a  line  beginning  at  a  given  point,  as  a  stake  driven 
in  the  ground,  and  parallel  to  a  curb,  fence  or  side  of  a  building, 
(a)  Use  the  method  described  in  194,  Exercise  4. 
8 


114  PLANE    GEOMETRY 

(6)   Use  the  method  described  in  194,  Exercise  6. 
(c)   The  most  accurate  method  is  suggested  in  68,  Exercise  11. 
This  may  be  used  as  a  check  on  the  accuracy  of  methods  (a) 
and  (6). 

2.  Measure  the  angles  of  a  four-sided  field  (or  at  four  stakes  driven 
in  the  ground),  in  order  to  verify  the  principle  of  187.  Distribute 
the  total  error  as  in  194,  Exercise  21. 

197.  STANDARD  FORM  SHEET  FOR  RECORDING  FIELD 
WORK. 

FIELD  WORK 

DEPARTMENT  OF  MATHEMATICS SCHOOL 

To  lay  out  a  line  beginning  at  a  given  point,  parallel  to  a 
given  line: 

Geometric  Principles  Employed. — (1)  If  the  diagonals  of  a 
quadrilateral  bisect  each  other,  the  figure  is  a  parallelogram 
(178). 

(2)  A  line  which  bisects  two  sides  of  a  triangle  is  parallel 
to  the  third  side  (180). 

(3)  Two  lines  which  are  cut  by  a  transversal  making  a 
pair  of  alternate-interior  angles  equal,  are  parallel  (72). 


\K   „---.'' 

/    -.-.-*""  V 


AB  is  a  curb,  fence,  side  of  a  building,  etc.;  C  is  a  point  on  the  required 

parallel  line. 

METHOD   (a) : 

(1)  Drive  stakes  at  D  and  E  on  line  AB. 

(2)  Measure  EC;  divide  by  2;  measure  to  middle 
point  F. 

(3)  Measure  DF;  prolong  until  FG  =  DF. 


POLYGONS  115 

METHOD  (6) : 

(1)  Measure  DC;  prolong  until  CH  =  DC. 

(2)  Measure  HE;  bisect  at  K. 
METHOD  (c). 

(1)  Measure   ADEC. 

(2)  Lay  out   Z.ECL  =    ^DEC. 

Points  C,  K,  G,  and  L  should  lie  in  the  same  straight  line. 


CHAPTER  VIII 
LOCI 

PRINCIPLES  DETERMINED  EXPERIMENTALLY 

198.  A  Locus  Defined. — A  line  may  be  considered  as 
being  made  up  of  an  indefinite  number  of  points.     When 
all  the  points  which  compose  a  line  possess  some  common 
property  of  position,  the  line  is  called  a  locus. 

A  locus  is  (in  general)  a  line  which  is  made  up  of  all  the 
points  which  satisfy  a  given  condition.  The  line  must 
contain  all  possible  points  which  satisfy  the  condition  and 
no  other  points.  A  point  which  moves  according  to  a 
fixed  law  generates  a  locus  or  part  of  a  locus. 

A  locus  may  also  be  a  group  of  lines,  a  line  and  a  point,  a 
series  of  detached  points,  a  surface,  etc. 

199.  Experiment  I. — To  find  the  locus  of  points  which 
are  equidistant  from  two  given  points. 

Draw  two  points  A  and  B.  To  find  a  point  of  the 
required  locus,  draw  two  arcs  with  centers  at  A  and  B 
respectively,  and  with  any  equal  radii,  which  intersect 
each  other.  Obtain  about  8  or  10  points  in  this  way. 
Draw  the  line  which  contains  all  the  determined  points. 
This  line  is  the  required  locus,  since  if  all  possible  points 
equidistant  from  points  A  and  B  were  found,  the  points 
themselves  would  form  the  line. 

RESULT. — The  locus  of  points  which  are  equidistant 
from  two  given  points  is  the  perpendicular-bisector  of  the 
sect  joining  the  given  points. 

200.  Experiment  II. — To  find  the  locus  of  points  which 
are  at  a  fixed  (or  constant)  distance  from  a  given  point. 

Mark  the  given  point  A  and  draw  the  given  fixed  distance 
BC.  Construct  the  locus  by  finding  a  number  of  detached 

116 


LOCI  117 

points  and  then  drawing  the  line  which  an  indefinite  number 
of  such  points  would  form.  B  c 

RESULT. — The  locus  of  points  which  H — H 

are  at  a  fixed  distance  from  a  given  point 
is  a  circle  whose  center  is  the  given  point 
and  whose  radius  is  the  fixed  distance. 

201.  Experiment    III.— To    find    the 
locus   of   points  which   are  at  a  fixed 
distance  from  a  given  straight  line. 

Help. — The  locus  is  composed  of  two     ^ 
lines. 

202.  Experiment  IV. — To  find  the  locus  of  points  within 
an  angle  which  are  equidistant  from  the  sides  of  the'angle. 

The  perpendicular  dis- 
tances from  each  point  of 
the  locus  to  the  sides  of 
the  angle  are  equal.    Such 
a  point  may  be  found  by 
trial   by   using  the  com- 
passes   to    measure    the  ~£ 
equal  distances,  and  plac-  A 
ing  them  so  that  they  measure  a  perpendicular  to  the  side 
of  the  angle,  the  perpendicular  being  carefully  judged,  or 
a  right  triangle  being  used  for  accuracy. 

203.  EXPERIMENT  V. — To  FIND  THE  Locus  OF  POINTS 
WHICH  ARE  TWICE  AS  FAR  FROM  ONE  OF  Two  GIVEN  POINTS 

AS  FROM  THE  OTHER. 

Mark  two  points  A  and  B. 

To  find  a  point  of  the  locus,  draw  any  straight  line  M N; 
mark  a  point  DI,  on  MN,  and  measure  DiDz  =  MD\. 
Draw  an  arc  with  center  at  A  and  radius  =  MD\;  and  a 
second  arc  with  center  at  B  and  radius  =  MDz;  the  points 
of  intersection  of  these  arcs,  D  and  D',  are  points  of  the 
required  locus. 

Repeat  with  radii  MEi,  and  ME2,  which  determine  the 
points  E  and  Ef  of  the  locus.  Find  enough  points  to 
completely  determine  the  locus,  which  will  be  found  to  be  a 


118  PLANE    GEOMETRY 

circle.     A  second  circle  may  be  found  if  the  relative  dis- 
tances of  the  points  of  the  locus  from  A  and  B  are  reversed. 


M          F,D,  E,  F2  D2       Ez  N 


B 

The  result  may  be  stated  if  desired,  by  stating  (1)  the 
form  of  the  locus,  (2)  the  position  of  the  center  of  the  circle, 
(3)  the  radius  of  the  circle.  (2)  and  (3)  are  stated  in 
terms  of  the  sect  AB. 

204.  REVIEW  EXERCISES 

Construct  the  following  loci  by  finding  only  sufficient  points  in 
each  to  determine  the  required  locus. 

1.  The  locus  of  points  which  are  equidistant  from  the  ends  of  a  sect. 

2.  The  locus  of  points  which  are  equidistant  from  two  parallel 
straight  lines. 

3.  The  locus  of  points  which  are  equidistant  from  two  intersecting 
straight  lines. 

4.  The  locus  of  the  centers  of  circles  which  contain  two  given  points. 

5.  The  locus  of  the  centers  of  circles  of  fixed  radius  tangent  to  a 
given  straight  line. 

6.  The  locus  of  the  centers  of  circles  of  fixed  radius  which  contain 
a  given  point. 

7.  The  locus  of  points  which  are  equidistant  from  two  concentric 
circles. 

205.  Algebraic  Equations  Considered  as  Loci. — Pupils 
who  have  studied  graphs  and  plotting  points  from  their 
coordinates,  will  understand  the  following  geometric  inter- 
pretation of  algebraic  equations  in  two  variables, 


LOCI 


119 


-r 


EXERCISES 

1.  Plot  a  number  of  points  for  which  x  =  y.     If  all  possible  points 
were  plotted,  they  would  make  up  a  line  bisecting  one  of  the 
pairs  of  vertical  angles  formed 

by  the  coordinate  axes,  x  — 
y  =  0  is  therefore  the  equa- 
tion of  this  line,  or  locus. 

2.  What  is  the  equation  of  the  *"' 
locus    of    points    equidistant  « 
from  the  sides  of  the  other                -10  -5 

pair  of  vertical  angles  formed  /     +5  */0 

by  the  coordinate  axes?  /^ — 1-5" 

3.  Combine  these  equations  into  > 
a     single    equation    with    a             ^/ 
double   (±)  sign.      Of  what 

may  this  equation  be  con- 
sidered the  locus? 

4.  Write  the  equation  of  the  locus  of  points  which  are  at  a  distance 
+5  or  —5  from  the  axis  of  abscissas. 

5.  Plot  the  locus  x  —  2y  =  12;  that  is,  the  locus  of  points  of  which 
the  abscissa  exceeds  twice  the  ordinate  by  12.     It  is  preferable 
to  use  squared  paper  to  facilitate  plotting. 

6.  Plot  the  locus  z2  +  y2  =  100.     To  what  locus  of  this  chapter 
does  the  graph  correspond? 

206.  Constructions  Depending  upon  Loci. — The  location 
of  a  point  which  fulfils  two  independent  given  conditions 
may  be  determined  by  drawing  the  two  loci  which  contain 
the  required  point.  The  intersection  of  these  loci  determines 
the  point. 

It  often  happens  that  more  than  one  point  fulfils  the 
given  conditions.  It  is  understood  that  all  such  points  are 
required. 

Certain  peculiarities  or  special  cases  of  the  problem  which 
may  occur  should  also  be  noted. 


EXERCISES 

1.  Find  a  point  equidistant  from  two  given  points  and  also  equi- 
distant from  two  given  parallel  lines. 

Let  A  and  B  be  the  given  points,  and  CD  and  EF  be  the 
given  parallels. 


120 


PLANE    GEOMETRY 


The  required  point,  in  order  to  be  equidistant  from  points  A  and 
B,  lies  on  the  perpendicular-bisector  of  the  sect  AB;  and  in  order 
to  be  equidistant  from  the  parallels  CD  and  EF,  it  lies  on  the 
line  PR,  parallel  to  them  and 
equidistant  from  them.  The 
point  0,  which  is  the  intersec- 
tion of  the  two  loci  is  therefore 
the  required  point. 
Special  Cases.— (a)  If  PR  is 
parallel  to  MN,  there  are  two 
points  at  infinity. 
(6)  If  PR  coincides  with  MN, 
every  point  of  MN  is  a  required 
point. 

(c)  There  is  no  impossible  case; 
unless  we  prefer  to  designate 
Case  (a)  as  impossible. 

2.  Find  a  point  equidistant  from 

two  given  points  and  also  equi-distant  from  the  sides  of  a 
given  angle. 

Help. — The  sides  of  the  angle  may,  if  necessary,  be  extended 
through  the  vertex  and  the  bisector  of  the  angle  may  therefore 
also  be  extended  through  the  vertex. 

3.  Find  a  point  equidistant  from  two  given  points  and  situated 
upon  a  given  straight  line. 

4.  Find  a  point  equidistant  from  two  given  points  and  also  at  a 
given  (fixed)  distance  from  a  given  straight  line. 

Help. — There  are  in  general  two  points  which  fulfil  these  condi- 
tions, and  several  special  cases. 

5.  Find  a  point  equidistant  from  two  given  points  and  situated 
on  a  given  circumference. 

Help. — There  are  in  general  two  points.  Special  cases  occur 
(a)  where  there  is  one  point,  and  (6)  where  there  is  no  point. 

6.  Find  a  point  equidistant  from  the  sides  of  an  angle  and  situated 
upon  a  given  straight  line. 

7.  Find  a  point  equidistant  from  two  given  points  and  at  a  given 
distance  from  a  third  given  point. 

8.  Find  a  point  equidistant  from  three  given  points. 

Help. — Find  a  point  equidistant  from  points  (1)  and  (2),  and 
also  equidistant  from  points  (2)  and  (3). 

9.  Find  a  point  within  a  triangle  and  three  points  outside  the 
triangle,  which  are  equidistant  from  the  three  sides  of  the  tri- 
angle (or  sides  extended). 

10.  Find  a  point  equidistant  from  two  intersecting  straight  lines 
and  situated  upon  a  third  given  straight  line. 


LOCI 


121 


11.  Find  a  point  equidistant  from  two  intersecting  straight  lines 
and  situated  upon  a  given  circumference. 

Help. — Besides  the  general  case,  there  are  five  special  cases. 

12.  A  pirate  buried  his  treasure  250  feet  from  an  oak  tree  O,  and 
70    feet    from    the    line    marked    by    two 

poplars  P  and  P'.  A  sailor  who  discovered 
the  pirate's  chart  and  directions  failed  to 
locate  the  treasure.  What  may  have  been 
his  error? 

13.  A  tree  is  20  feet  from  a  straight  fence.     A 
gardener  directs  his   assistant   to   plant   a 
bush   30  feet  from   both  the  tree  and  the 

fence.     Are  the  directions  sufficient?     For  what  case  would 
they  be  sufficient?     Draw  a  sketch. 

14.  Find  a  point  equidistant  from  two  perpendicular  axes,  that  is, 
on  the  locus  x  ±  y  =  0;  and  also  at  a  distance  of  5  units  from 
the  vertical  axis,  that  is,  on  the  locus  x  =   ±  5. 

Help. — There  are  four  points. 

APPLICATIONS 

207.  SURVEYING. — A  point  equidistant  from  three  given 
points  may  be  located  by  the  method  of  206,  Exercise  8. 

The  lines  are  bisected  by  measurement,  and  the  perpen- 
diculars are  erected  either  by  measuring  a  right  angle  with 
a  transit  or  by  some  method  in  which  a  tape  is  used. 

208.  Higher    Mathematics. — Coordinate    geometry,    de- 
scribed briefly  in  205,  forms  the  basis  of  calculus  and  of  a 
great  deal  of  applied  mathematics,  such  as  engineering, 
military   science   and    navigation.     An   understanding  of 
loci  is  therefore  a  requisite  for  such  professional  studies. 

209.  THE  ELLIPSE. — This  curve  is  the  locus  of  points  the 
sum  of  whose  distances  from  two  given 

points  is  a  constant. 

To  construct  an  ellipse:  (1)  draw  the 
axes  AB  and  CD  bisecting  each  other 
at  right  angles;  (2)  draw  two  arcs  with 
center  at  C  and  radii  equal  to  AG, 
intersecting  AB  at  points  E  and  F,  which  are  called  the 
foci;  (3)  tie  two  knots  in  a  string  at  a  distance  apart 
equal  to  AB;  fix  the  knots  with  pins  at  points  E  and  F, 


122  PLANE    GEOMETRY 

and  slip  a  pencil  around  the  line  AHKCB  keeping  the 
string  taut. 

The  locus  is  correctly  described  since  EA  +  AF  =  EH 
+  HF  =  EC  +  CF  =  etc.  =  a  constant,  which  =  AB. 

EXERCISES 

1.  Construct  an  ellipse  whose  axes  are  5  and  3  inches  respectively. 

2.  Lay  out  an  elliptical  flower  bed  24  feet  by  12  feet. 

3.  Draw  an  elliptical  window  with  symmetrical  colored  glass  design. 

4.  Draw  an  elliptical  picture-frame,  platter,  or  embroidered  doily 
or  table-cover. 

5.  Observe  occasional  elliptical  outlines  in  architecture  as  arches, 
ceilings,  column  bases,  windows.     These  are,  in  general,  more 
pleasing  than  circular  arcs. 


CHAPTER  IX 
LOCI 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

210.  The  Method  Employed. — After  it  has  been  deter- 
mined experimentally  that  a  certain  line  (or  lines)  is  made 
up  of  points  having  some  given  or  assigned  property;  it  is 
necessary  to  show,  in  order  to  explain  this  principle:  (a) 
that   every   point   in  the   determined   line   possesses   the 
assigned  property;  and  (6)  that  every  possible  point  which 
possesses  the  assigned  property  is  in  the  determined  line. 

If  only  property  (a)  were  known  about  a  determined  line, 
the  line  still  might  be  only  a  part  of  the  locus.  If  only 
property  (6)  were  known  about  the  determined  line,  the 
entire  line  might  be  more  than  the  required  locus.  If  both 
properties  (a)  and  (6)  belong  to  the  line,  then  the  determined 
line  is  the  complete  locus. 

Only  two  of  the  loci  determined  experimentally  in 
Chapter  VIII  are  important  in  advanced  analysis.  The 
four  concurrent  line  properties  of  a  triangle  (111)  which 
have  not  as  yet  been  analyzed,  are  explained  from  the 
properties  of  loci  (213-216). 

211.  Theorem  I. — The  locus  of  points  equidistant  from 
the  ends  of  a  sect  is  the  perpendicular-bisector  of  the  sect. 

Part  (a). — Any  point  on  the  perpendicular-bisector  of  a 
sect  is  equidistant  from  the  ends  of  the  sect. 


c 
123 


124 


PLANE    GEOMETRY 


Hypothesis. — Sect  AB  with  perpendicular-bisector  CD, 
and  a  point  E  on  CD. 

Conclusion. — Point  E  is  equidistant  from  points  A  and  B; 
i.e.,  AE  =  BE. 

Analysis.— (1)  AECA  ^  ECB;  (2)  .'.  AE  =  BE.  Write 
out  the  complete  analysis  in  standard  form. 

Part  (6). — Any  point  which  is  equidistant  from  the  ends 
of  a  sect  is  on  the  perpendicular-bisector  of  the  sect. 


Hypothesis. — Sect  AB  with  perpendicular-bisector  CD, 
and  a  point  F  equidistant  from  A  and  B;  i.e.,  AF  =  BF. 

Conclusion. — Point  F  is  on  CD. 

Analysis.— (1)  Draw  CF;—(2)  AFC  A  *  AFCB;—(3)  .'. 

CF  _L  AB;—  (4)  .'.  CF  coincides  with  CD  and  F  is  on  CD. 

Write  out  complete  analysis  in  standard  form. 

Corollaries.  1. — Two  points  equidistant  from  the  ends 
of  a  sect  determine  the  perpendicular-bisector  of  the  sect. 

Helps. — (1)  Each  point  is  on  the  perpendicular-bisector; 
why?  (2)  .'.  the  line  joining  the  two  points  coincides  with 
the  perpendicular-bisector. 

2.  //  a  perpendicular  to  a  sect  contains  a  point  that  is 
equidistant  from  the  ends  of  the  sect,  it  bisects  the  sect. 

Helps. — This  may  be  analyzed  by  principles  of  congruent 
triangles,  or  of  isosceles  triangles,  or  as  follows: — (1)  the 
perpendicular-bisector  of  the  sect  contains  the  given  point; 
why? — (2)  .'.  the  two  perpendiculars  coincide,  etc. 


LOCI 


125 


212.  Theorem  II. — The  locus  of  points  within  an  angle 
equidistant  from  the  sides  is  the  bisector  of  the  angle. 
State  Part  (a). 


State  the  hypothesis,  and  the  conclusion  (EF  =  EG) 
State  Part  (6). 


State  hypothesis  (HK  =  HL);  and  the  conclusion. 
'  Helps.— (I)    Consider  &HLB  and  HKB;—(2)  if  x  =  y, 
BH  bisects  /.ABC;—  (3)   /.  BH  coincides  with  BD,  and 
point  H  is  on  BD. 

Corollary. — One  point  equidistant  from  the  sides  of  an 
angle  determines  with  the  vertex  the  bisector  of  the  angle. 

213.  Theorem  III. — The  three  perpendicular-bisectors  of 
the  sides  of  a  triangle  are  concurrent. 


Hypothesis. — A  ABC     with     perpendicular-bisectors     of 
sides,  DE,  FG  and  HK. 

Conclusion. — DE,  FG  and  HK  are  concurrent. 


126 


PLANE    GEOMETRY 


Analysis. 

STATEMENT 

1.  DE  and  FG  intersect  at  a  point  0 

2.  Pt.  0  is  equidistant  from  pts.  A 

and  B;  that  is,  AO  =  OB 

3.  Pt.  0  is  equidistant  from  pts.  B 

and  C;  that  is,  OB  =  OC 

4.  .'.  Pt.  0  is  equidistant  from  pts.  A 

and  C;  that  is,  OA  =  OC 

5.  .'.  HK  contains  pt.  0 


6. 


REASON 

143,  Theorem  (3) 
Locus  Theorem  I, 

Part  (a) 
Locus  Theorem  I, 

Part  (a) 
Axiom  1 

Locus  Theorem  I, 
Part  (6) 


,  FG  and  HK  are  concurrent 
Corollary. — Point  0  is  equidistant  from  the  three  vertices 
of  the  triangle. 

214.  Theorem  IV. — The  three  angle-bisectors  of  a  triangle 
are  concurrent. 

Helps. — (1)  Two  of  the  angle-bisectors  intersect  at  a 
point  0;  what  theorem?  (2)  point  0  is  in  the  third  angle- 
bisector. 

Write  out  the  complete  theorem  using  statements 
identical  with  those  of  Theorem  III,  except  that  point  0  is 
here  equidistant  from  the  sides  of  the  angles. 

Corollary. — Point  0  is  equidistant  from  the  three  sides  of 
the  triangle. 

215.  Theorem  V. — The  three  altitudes  of  a  triangle  are 
concurrent. 


Helps.— (1)  Draw  GH,  HK,  KG\\AB,  BC,  CA  respec- 
tively;—(2)  AD.LKH;  why?—  (3)  KACB  is  a  parallelogram; 
.'.  KA  =  BC;—(4)  also  AH  =  BC;  why?— (5)  /.  KA  = 


LOCI  127 

AH;—  (6)  .'.  AD  bisects  KH.—(7)  Similarly  BE  and  CF 
are  perpendicular-bisectors  of  the  sides  KG  and  GH  of 
AGHK. — (8)  Consider  now  that  A  ABC  is  removed,  leaving 
AGHK  and  the  three  perpendicular-bisectors  of  its  sides; 
they  are  concurrent;  why? 

Write  out  the  complete  analysis. 

216.  THEOREM  VI. — The  bisectors  of  each  two  exterior 
angles  of  a  triangle  and  of  the  non-adjacent  interior  angle  are 
concurrent. 


State  the  hypothesis  and  the  conclusion. 

Helps. — (1)  AD  and  BE  intersect  in  a  point  0; — (2)  which 
is  equidistant  from  CA  and  CB; — (3) .'.  CF  contains  point  0. 
Show  that  there  are  two  other  similar  points. 

State  a  corollary  similar  to  the  corollaries  of  Theorems 
III  and  IV. 

ADDITIONAL  THEOREMS 

217.  (1)  If  any  point  of  the  median  to  the  base  of  an 
isosceles  triangle  is  joined  to  the  ends  of  the  base,  another 
isosceles  triangle  is  formed. 

Help. — The  median  is  the  perpendicular-bisector  of  the 
base. 

(2)  The  middle  point  of  the  hypotenuse  of  a  right  triangle 
is  equidistant  from  the  three  vertices. 

Help. — Show  that  the  perpendicular-bisectors  of  two  of 
the  sides  of  the  triangle  are  concurrent  at  this  point. 

(3)  The  circumcenter,  incenter,  orthocenter  and  centroid 
of  an  equilateral  triangle  coincide. 

(4)  The  diagonals  of  a  quadrilateral  divide  it  into  four 


128  PLANE    GEOMETRY 

triangles   whose   circumcenters   are   at  the   vertices   of   a 
parallelogram. 

(5)  The  perpendicular-bisector  of  a  chord  of  a  circle 
contains  the  center  of  the  circle. 

(6)  The  line  of  centers  of  two  intersecting  circles  is  a 
perpendicular-bisector  of  the  chord  joining  their  points  of 
intersection. 

218.  REVIEW  EXERCISES 

1.  Sketch  the  locus  of  points  which  are  twice  as  far  from  one  of 
the  sides  of  an  angle  as  from  the  other:  (a)  when  the  points  lie 
within  the  angle;  (6)  when  they  lie  outside  of  the  angle. 

2.  Construct  by  points,  the  locus  of  the 
vertex  of  an  angle  of  constant   size, 
whose  sides   pass  through  two  given 
points. 

Help. — Use  an  angle  of  the  drawing     - 
triangle.     Place  the  triangle  in  differ- 
ent positions  such  that  the  sides  form- 
ing the  angle  contain  the  two  marked  points. 

3.  If  a  locus  is  known  to  be  a  straight  line,  how  many  points  are 
required  to  determine  it? 

4.  Construct  the  locus  of  the  centers  of  circles  which  are  tangent 
to  the  two  sides  of  an  angle. 

5.  Find  a  point  (or  points)  which  is  equidistant  from  two  given 
points  and  also  at  a  given  distance  from  a  given  circumference. 

6.  Find  a  point  which  is  equidistant  from  two  intersecting  straight 
lines  and  at  a  given  distance  from  their  point  of  intersection. 

7.  Plot  the  locus  xy  =  12,  using  both  plus  and  minus  values  of  the 
coordinates. 

8.  A  stone  wall  across  a  farm  runs  north  and  south;  a  second  wall 
crosses  it  east  to  west;  200  ft.  east  of  the  point  of  intersection  a 
third  wall  crosses  the  second  from  S.E.  to  N.W.  at  an  angle  of 
50°.     The  owner  of  the  farm  finds  in  an  old  paper:  "trap-door 
to  concealed  passage  leading  to  treasure  vault,  is  at  point  equally 
distant  from  three  walls."     Draw  map  and  locate  the  trap-door. 
In  how  many  possible  places  may  it  be  located? 

9.  The  first  mathematician  to  use  the  coordinate  method  of  plotting 
algebraic  equations  was  Rene  Descartes   (1596-1650).     Coor- 
dinate geometry  is  also  called  Cartesian  geometry  in  his  honor. 
Find  in  an  encyclopedia  some  facts  concerning  Descartes. 


CHAPTER  X 

THE    MEASUREMENT    OF    SECTS,    AND    SIMILAR 
TRIANGLES 

PKINCIPLES  DETERMINED  EXPERIMENTALLY 

219.  In  this   chapter  we   propose  to  investigate   more 
complex  relationships  of  sects  than  those  merely  of  equality 
or  inequality,  these  relationships  forming  an  entirely  new 
group  of  geometric  principles.     Some  of  the  more  evident 
of  these  principles  have  certainly  been  known  as  early  as 
2000  B.C.,  and  researches  in  this  field  were  made  by  Pytha- 
goras about  500  B.  C. 

220.  Measurement  of  Sects. — A  ratio  is  a  quotient  arising 
from  the  division  of  one  quantity  by  another  of  the  same 

kind;  as  r,  a  -r-  6,  a  :  b. 

a      c 
A  proportion  is  an  equality  of  ratios;  as  T  =  ^,  a  -r-  b  = 

c  -T-  d,  a  :  b  =  c  :  d. 

The  measurement  of  a  quantity  is  a  ratio  of  the  quantity 
measured,  to  the  unit  of  the  same  kind.  See  also  Chapter  I. 

The  ratio  of  two  quantities  of  the  same  kind  is  the  ratio 
of  their  measures  in  terms  of  a  common  unit. 

EXERCISES 

1.  What  units  are  generally  used  to  measure  sects  of  about  the 
length  of  this  page?  very  short  sects?  the  side  of  a  room?  geo- 
graphical distances?  astronomical  distances? 

2.  Measure  sects  a  and  b  with  the  unit  c.     What  is  the  ratio  of  £  ? 

b 


3.  Measure  the  edges  of  this  page  in  inches.     Draw  two  sects  less 
than  two  inches  whose  ratio  equals  the  ratio  of  the  two  edges. 
9  129 


130  PLANE    GEOMETRY 

221.  Properties  of  a  Proportion. — ,  =  ^,  or  a  :  b  =  c  :  d, 

represents  any  proportion. 

The  terms  are  the  numbers  a,  b,  c  and  d. 

The  means  are  the  second  and  third  terms,  6  and  c. 

The  extremes  are  the  first  and  fourth  terms,  a  and  d. 

The  four  numbers  (quantities,  values)  are  proportional. 

The  fourth  term  of  a  proportion  is  a  fourth  proportional 
of  the  three  preceding  terms. 

When  the  two  means  are  the  same,  either  is  a  mean 
proportional  of  (between)  the  first  and  fourth  terms. 

A  reciprocal  of  a  number  is  unity  divided  by  the  number. 

The  numbers  are  reciprocals,  as  x  and  -. 

x 

EXERCISES 

7       14 

1.  Name  the  means  and  extremes  of  the  proportions,  g  =  -g-; 

5  :  x  =  15  :  2;  x  :  x  +  1  =  6  :  8. 

2.  What  is  the  fourth  proportional  of  5,  8  and  25? 

3.  What  is  the  mean  proportional  of  2  and  8;  of  1  and  25?     Express 
the  proportions. 

4.  Write  four  proportional  numbers. 

5.  Write  a  series  of  three  equal  ratios  of  which  the  first  ratio  is  7,- 

6.  Write  as  many  simple  proportions  (two  ratios  in  each)  as  possi- 

a       c       e 
ble  from  the  ratio  series,  r  =  -;  =  j. 

7.  Combine  these  proportions   into   a  series,   v  =  -  and   r  =  -.. 

What  axiom  is  used? 

1       2 

8.  Write  the  reciprocal  of  5;  of  gjof  ^;  of  .25. 

222.  Division  of  a  Sect. — A  sect  is  divided  by  a  point,  or 
intersecting  line,  into  two  parts  or  segments. 

When  the  point  is  between  the  ends  of  the  sect  it  divides 
the  sect  internally;  when  the  point  is  in  the  sect  extended  it 
divides  the  sect  externally.  The  segments  extend  from  the 
point  of  division  to  each  end  of  the  sect. 

A  sect  is  divided  in  a  given  ratio  when  the  ratio  of  the 
segments  equals  the  given  ratio. 

Two  sects  are  divided  proportionally  when  the  four  seg- 
ments are  proportional. 


MEASUREMENT   OF   SECTS,    AND   SIMILAR   TRIANGLES  131 

EXERCISES 

1.  The  sect  AB  is  divided  internally  at  C  and  externally  at  D. 


B 


Read  the  segments  of  AB  formed  by  each  point  of  division. 

2.  Draw  a  sect  3  inches  long.     Mark  a  point  which  divides  it  in- 
ternally in  the  ratio  of  1  :  2.     Mark  a  point  which  divides  it 

o 

internally  hi  the  ratio  of  „• 

3.  Divide  the  sect  of  Exercise  2  externally  in  the  same  ratios. 
Help. — Let  x  =  segment  DB;  then  x  +  3  =  segment  DA;  and 

x       _  1 
x  +  3  ~  2* 

4.  Draw  sects  of  3  and  4  inches.     Divide  them  proportionally  so 

that  a  segment  of  the  3-inch  sect  =  „•  inch. 

5.  A  mile  is  divided  into  segments  of  2000  and  3280  feet.     Divide 
an  inch  proportionally,  the  segments  being  calculated  to  two 
decimals. 

6.  A  sect  is  said  to  be  divided  harmonically  when  it  is  divided  by 
two  points,  one  internal  and  the  other  external,  so  that  the  ratio 
of  the  segments  formed  by  one  point  of  division  equals  the  ratio 
of  the  segments  formed  by  the  other  point. 

Divide  a  sect  AB  =  2  inches  internally  at  C  so  that  EC  =  .5", 
and  harmonically  at  D.     Calculate  DB. 
„.        DA       CA 
Help'~~DB  =  OB' 

223.  Necessary  Accuracy  of  Experimental  Work. — In  the 
experimental  work  of  this  chapter,  measure  sects  to  the 
nearest  hundredth  of  an  inch  and  angles  to  the  nearest 
quarter  degree.     Drawings  must  be  made  very  accurate  in 
order  to  obtain  reliable  results. 

224.  Experiment  I. — The  ratio  of  two  sects. 

Draw  two  sects  of  different  lengths,  each  of  several 
inches. 

(a)  Find  their  measures  in  inches.  Calculate  the  ratio 
of  their  lengths  as  a  decimal,  to  three  places. 

(6)  Find  their  measures  in  centimeters  or  in  any  arbi- 
trary unit  as  H H  Calculate  their  ratio. 

Observe  the  relation  between  the  ratios. 


132  PLANE    GEOMETRY 

RESULT. — The  ratio  of  two  sects  is  independent  of  the 
unit  of  measurement. 

225.  Experiment  II. — A  line  parallel  to  a  side  of  a 
triangle. 

x   H ff/ 


Draw  any  oblique  scalene  triangle  ABC.  Draw  a  line 
parallel  to  a  side  of  the  triangle,  intersecting  the  other  two 
sides.  Measure  the  segments  into  which  the  two  sides  are 

AD 
divided  by  the  parallel  line.     Calculate  the  ratios  ^5  and 

AE 

7^77.  each  to  three  decimals.     Compare  these  ratios.     Re- 

JtLL 

peat  with  other  parallel  lines,  as  FG,  HK,  MN 

RESULT. — A  line  parallel  to  a  side  of  a  triangle  divides 
the  other  two  sides  proportionally. 

EXERCISES 

1.  The  sides  of  AABC  are  AB  =  2",  AC  =  3",  BC  =  2.4".     If 
DE  cuts  AB  so  that  AD  =  1.5",  calculate  AE. 

2.  In  the  triangle  of  Exercise  1,  MN  cuts  AC  so  that  NC  =  2.2"; 
calculate  MB. 

3.  In  the  same  triangle,  FG  cuts  AB  prolonged  so  that  BF  =  0.5; 
calculate  CG. 

4.  In  the  same  triangle  AH  =  |  AC;  calculate  AK. 

5.  If  point  D  is  the  middle  of  AB,  to  what  previous  simpler  prin- 
ciple does  this  principle  reduce? 

6.  The  sides  of  a  triangle  are  100  ft.,  250  ft.,  280  ft.     A  line  parallel 
to  the  280  ft.  side  cuts  the  250  ft,  side  60  ft.  from  its  intersection 
with  the  100  ft.  side.     At  what  point  does  this  line  cut  the  100  ft. 
side?     From  the  same  point  in  the  250  ft.  side,  a  line  runs  par- 


MEASUREMENT    OF   SECTS,    AND    SIMILAR   TRIANGLES  133 

allel  to  the  100  ft,  side.     At  what  point  does  it  cut  the  280  ft. 
side? 

7.  Two  sides  of  a  parallelogram  are  10"  and  15",  and  a  diagonal  is 
18".  At  a  point  in  this  diagonal,  3  inches  from  one  end,  lines 
are  drawn  parallel  respectively  to  the  four  sides  of  the  parallelo- 
gram. Calculate  the  points  at  which  these  lines  intersect  the 
sides. 

226.  Experiment   III. — The    position    of    a    line  which 
divides  two  sides  of  a  triangle  proportionally. 


Draw  any  triangle  ABC;  mark  a  point  D  on  AB.     Meas- 
ure AD,  AB  and  AC.     Calculate  the  position  of  a  point  E 

on  AC  such   that  T7*=T7?'     Measure   AE  and   mark 

point  E;  draw  DE.     Observe  the  position  of  line  DE. 

RESULT. — A  line  which  divides  two  sides  of  a  triangle 
proportionally,  etc. 

EXERCISES 

1.  The  sides  of  a  triangle  are  4",  6",  8".     A  line  cuts  the  4"  and  6" 
sides  at  points  1"  and  1.5"  respectively  from  their  point  of  inter- 
section.    Is  this  line  parallel  to  the  8"  side? 

2.  Two  adjacent  sides  of  a  rectangular  room  are  10  ft.  and  15  ft. 
A  line  joins  two  points  on  the  sides  of  the  room,  measured 
respectively  3  ft.  6  in.  and  5  ft.  3  in.  from  a  corner  of  the  room. 
Is  this  line  exactly  or  approximately  parallel  to  a  diagonal  of 
the  room. 

227.  Similar  triangles  are  defined  in  92. 

228.  Experiment  IV. — The  relation  of  the  sides  of  similar 
triangles. 


134  PLANE    GEOMETRY 

Draw  any  oblique  scalene  triangle;  measure  the  three 
angles  with  a  protractor.  Draw  a  second  triangle  of  a 
different  size  whose  angles  are  equal  to  the  angles  of  the 
first  triangle.  Measure  all  the  sides  of  both  triangles. 

Calculate  the  ratio  of  each  pair  of  homologous  sides,  — ,  ^,  —f, 

to  three  decimals.     Observe  the  relation   between   these 
ratios.     State  result. 

EXERCISES 

1.  The  sides  of  a  triangle  are  3",  5",  5.4".     The  side  of  a  similar 
triangle,  homologous  to  the  3"  side,  is  6".     Calculate  the  other 
sides  of  the  second  triangle. 

2.  The  sides  of  a  AXYZ  are  x  =  27",  y  =  31",  z  =  22".     The 
side  y'  of  a  similar  AX'Y'Z'  is  42".     Calculate  sides  xf  and  z'. 

229.  Experiment  V. — Triangles  whose   sides  are  pro- 
portional. 

Draw  any  triangle;  measure  the  three  sides.  Multiply 
each  side  by  the  same  number,  as  .0.73,  or  1.5,  etc.  Con- 
struct a  second  triangle  from  a  set  of  three  sides  thus 
obtained.  Measure  the  three  angles  of  both  triangles. 
Observe  the  kind  of  triangles  and  state  result. 

EXERCISES 

1.  The  sides  of  two  triangles  are  respectively  7",  5",  3.4",  and 
21",  15",  10.2".     Are  the  triangles  similar? 

2.  The  sides  of  two  triangles  are  respectively  4.5  ft.,  5.8  ft.,  7.1  ft., 
and  5.4  ft.,  6.96  ft.,  9.94  ft.     Are  the  triangles  similar? 

3.  If  a  triangle  is  drawn  with  sides  a  =  6",  b  =  8",  c  =  10",  the 
angles  will  be  found  to  measure  approximately,  A  =  37°,  B  = 
53°,  C  =  90°.     Obtain  the  values  of  the  angles  of  a  triangle 
DEF,  of  which  the  sides  are  d  =  15",  e  =  25",  /  =  20",  without 
constructing  the  triangle. 

230.  Experiment  VI. — Triangles   with   two   sides   pro- 
portional and  the  included  angles  equal. 

Draw  any  triangle;  measure  two  sides  and  the  included 
angle.  Construct  the  second  triangle.  Measure  the  other 
two  angles  and  the  third  sides  of  both  triangles.  Observe 
(a)  the  kind  of  triangles  obtained,  and  (b)  the  ratio  of  the 
third  sides. 


MEASUREMENT    OF   SECTS,    AND    SIMILAR   TRIANGLES  135 

EXERCISES 

1.  If  two  triangles  are  constructed  from  the  parts,  respectively, 
a  =  5",  b  =  7",  C  =  40°,  and  a'  =  10",  V  =  14",  C"  =  40°; 
are  the  triangles  similar? 

2.  If   A  ABC  is  constructed  from  the  values  a  =  10",  6  =  30", 
C  =  45°,  the  remaining  parts  will  be  found  to  be  approximately, 

c  =  24",  A  =  17°,  B  =  118°.  In  a  ADEF,  d  =  15",  /  =  45", 
E  =  45°.  Determine  without  constructing  the  triangle,  the 
values  of  parts  e,  D  and  F. 

231.  Projection. — The  projection  of  a  point  on  a  line  is  the 
foot  of  the  perpendicular  from  the  point  to  the  line.  The 
given  line  is  the  line  of  projection. 

The  projection  of  a  sect  on  a  given  line  is  the  sect  included 
by  the  projections  of  its  end  points. 

EXERCISES 

1.  Draw  a  point  A  and  a  line  BC;  pro- 
ject A  upon  BC.     Draw  a  sect  DE; 
project  DE  upon  BC.    Name  the 
line  of  projection. 

2.  When  does  the  projection  of  a  sect 
become  a  point?     When  is  it  equal 

to  the  sect?     Is  it  ever  greater  than  B  ~ 
the  sect? 

3.  Draw  a  sect  AB  intersecting  a  line  CD  at  A.     Construct  the 
projection  of  AB  on  CD. 

4.  Draw  two  intersecting  sects;   con- 
struct  the  projection  of   each  on 
the  other. 

5.  Draw  a  triangle:  project  each  two 
sides  on  the  third  side. 

6.  In  what  kind  of  triangle  do  the  projections  of  any  two  sides  upon 
the  other  side  fall  within  the  side  upon  which  the  projection  is 
made?    In  what  kind  of  triangle  do  the  projections  of  some 
sides  fall  on  a  side  prolonged?     In  what  kind  of  triangle  is  the 
projection  of  one  or  more  sides  a  point? 

7.  In  every  triangle,  how  do  the  projections  of  any  two  sides  upon 
the  third  side  compare  with  the  length  of  the  third  side? 

8.  Construct  a  sect  AB  =  3",  at  such  an  angle  to  line  CD  that  its 
projection  on  CD  =  2.2". 


136  PLANE    GEOMETRY 

232.  EXPERIMENT  VII. — THE  RELATION  BETWEEN  THE 
SEGMENTS  OF  A  SECT  AND  THE  PROJECTIONS  OF  THE  SEGMENTS. 

Draw  a  sect;  divide  it  into  two  or  more  unequal  segments. 
Draw  any  other  indefinite  line,  not  parallel  to  the  sect. 
Project  the  segments  of  the  sect  upon  the  other  line. 
Measure  the  segments  and  their  projections.  Observe 
how  certain  ratios  compare.  State  result. 

EXERCISES 

1.  A  sect,  AB,  3"  long  is  divided  at  point  C  into  segments  AC  = 
1.2"  and  CB  =  1.8".     The  segments  are  projected  on  a  line 
DE  such   that  the   projection  of  AC  =  0.8".     Calculate   the 
projection  of  segment  CB. 

2.  The  projections  of  the  segments  AC  and  CB  of  a  sect  AB  are 
3.5"  and  6.2".     Sect  AB  =  12".     Calculate  the  segments  of 
AB. 

233.  Experiment  VIII.— The  relation  between  the  sides 
of  a  right  triangle. 

Draw  one  or  more  right  triangles;  measure  the  sides. 
Square  the  value  of  each  side.  Observe  the  relation  be- 
tween the  squares  of  the  three  sides  of  each  triangle. 
State  result. 

EXERCISES 

1.  The  perpendicular  sides  of  a  right  triangle  are  3  and  4  inches. 
Calculate  the  hypotenuse. 

2.  The  hypotenuse  and  base  of  a  right  triangle  are  respectively 
13  and  5  inches.     Calculate  the  altitude. 

3.  May  the  lengths  8,  15,  17  inches  be  taken  as  the  sides  of  a  right 
triangle?     Why? 

234.  Experiment  IX. — The  condition  for  similar  polygons. 
Draw  a  trapezium  (a  polygon  of  four  sides). 

(a)  Draw  a  polygon  having  its  angles  equal,  each  to 
each,  to  those  of  the  given  polygon,  but  its  sides  in  no 
constant  ratio  to  the  sides  of  the  given  polygon. 

(b)  Draw  a  polygon  having  its  sides  proportional  to  the 
sides  of  the  given  polygon,  but  its  angles  in  no  relation  to 
'those  of  the  given  polygon. 


MEASUREMENT    OF    SECTS,    AND    SIMILAR   TRIANGLES  137 

(c)  Draw  a  polygon  having  its  angles  equal,  each  to 
each,  to  the  angles,  and  its  sides  proportional  to  the  sides, 
of  the  given  polygon. 

Observe  the  following  principles : 

(a)  May  two  polygons  be  mutually  equiangular  without 
being  similar? 

(6)  May  two  polygons  have  their  sides  proportional 
without  being  similar? 

(c)  State  the  complete  condition  for  similar  polygons  of 
more  than  three  sides. 

EXERCISES 

1.  A  rectangle  has  adjacent  sides  of  4"  and  2".     A  second  rectangle 
has  homologous  sides  of  6"  and  3",     Are  the  rectangles  similar? 

2.  The  sides  of  two  quadrilaterals  are  respectively,  3",  5",  4.8",  2"; 
and  (in  the  same  order)  6",  10",  9.6",  4".     Are  they  similar? 

3.  The  sides  of  one  of  two  similar  pentagons  are  7",  10",  5",  8",  6". 
A  side  of  the  second  pentagon,  homologous  to  the  10"  side  of 
the  first,  is  12".     Calculate  the  other  sides  of  the  second  pentagon. 

235.  Abbreviation. — Similar,  is  similar  to,  ~. 

APPLICATIONS 

236.  Forestry. — The  following  methods  of  determining 
the  heights  of  trees  with  sufficient  accuracy,  are  described 
in  a  bulletin  issued  by  the  U.  S.  Department  of  Agriculture. 

(1)  The  Shadow  Method. — Thales  employed  this  method 
to  find  the  height  of  the  great  pyramids  of  Egypt,  to  the 
admiration  of  native  scholars. 


EXERCISES 

1.  Are  the  sun's  rays,  AC  and  DF,  substantially  parallel? 

2.  Prove  that  the  triangles,  formed  by  tree,  shadow  and  sun's  ray 


138 


PLANE   GEOMETRY 


is  similar  to  that  formed  by  the  vertical  pole,  shadow  of  pole 
and  sun's  ray. 

3.  The  shadow  of  a  tree  is  measured  130  feet  at  the  same  time 
that  the  shadow  of  a  12-foot  pole  is  15  feet.     Calculate  the 
height  of  the  tree. 

4.  May  this  method  be  used  to  find  the  height  of  a  flagpole,  build- 
ing, stack,  etc. 

5.  Will  the  two  triangles  be  similar  if  the  ground  is  sloping?     If 
the  tree  is  leaning? 

6.  Find  a  few  facts  concerning  Thales. 

(2)  The  Graduated  Rod  Method. 

(a)  The  observer  standing. 
(6)  The  observer  lying  down. 


EXERCISES 

7.  Explain  both  methods  as  illustrated. 

8.  Place  letters  on  the  figures  and  read  the  similar  triangles. 

9.  What  measurements  are  taken  in  each  method? 

10.  Assume  reasonable  measurements  and  calculate  the  heights  of 
the  trees  by  both  methods. 

11.  Is  this  method  applicable  if  the  ground  is  sloping? 

(3)  The  Mirror  Method. 


EXERCISES 

12.  Explain  the  method  illustrated. 

13.  Where  are  the  similar  triangles? 

14.  Assume  reasonable  measurements  and  calculate  the  height  of 
the  tree. 

15.  Is  this  method  applicable  on  sloping  ground  if  the  mirror  is 
leveled? 


MEASUREMENT   OF   SECTS,    AND   SIMILAR   TRIANGLES  139 

(4)  Faustmann's  Hypsometer. — A  simplified  form  of  this 
instrument  is  shown.     The  instrument  is  sighted  to  the  top 


0      ' 


of  the  tree  and  the  reading  is  taken  at  F  on  the  scale  EF. 
It  is  necessary  to  measure  AB,  which  may  be  done  by 
pacing,  and  to  measure  or  estimate  BG.  This  may  be  done 
by  sighting  through  a  hand  level  and  noting  the  point  B 
on  the  level  line  AB,  and  then  measuring  from  B  to  the 
ground.  After  calculating  CB,  the  distance  BG  is  added 
to  find  the  total  height  of  the  tree. 

In  the  instrument  as  usually  made,  the  point  of  attach- 
ment D  of  the  plumb-line  is  on  a  movable  scale,  so  that  the 
distance  AB  in  feet  may  be  set  off  on  the  scale  equal  to  ED. 
The  graduations  on  scale  EEf  are  made  in  the  same  units  as 
employed  on  scale  ED,  and  the  height  of  the  tree  in  feet 
is  therefore  equal  to  the  reading  of  the  scale  distance  EF. 
A  narrow  adjustable  mirror  hinged  at  point  E  is  provided, 
and  the  observer  may  thus  read  the  scale  at  F  while  sighting 
the  instrument  on  C,  the  figures  of  the  scale  EE'  being 
reversed. 

EXERCISES 

16.  Name  the  two  similar  triangles.     Prove  that  they  are  similar. 

17.  If  DE  =  5",  EF  =  6",  AB  =  40',  BG  =  7';  calculate  CG. 

18.  If  BG  =  8.5  feet,  AB  =  120  feet;  and  if  ED  is  set  at  120  units 
and  EF  reads  95  units;  what  is  the  value  of  BC  in  feet,  and 
what  is  the  height  of  the  tree? 

237.  Surveying. — The  principles  of  similar  triangles  afford 
some  useful  methods  of  determining  distances  which  cannot 
be  measured  directly,  and  of  laying  out  perpendiculars. 


140 


PLANE    GEOMETRY 


The  Distance  Across  a  River. — The  distance  AB  is 
required.  Turn  a  right  angle  at  A;  measure  any  distance 
AC;  turn  a  right  angle  at  C; 
measure  any  distance  CD.  Set 
a  stake  at  point  E  on  both  lines 
DJSandAC. 


EXERCISES 

1.  Name  the  similar  triangles  of 
the   figure.      Prove   that   they 
are  similar. 

2.  What  proportion  can  be  stated  whose  solution  will  give  the 
distance  AB? 

3.  What  additional  measurement,  not  mentioned  above,  must  be 
made,  in  order  to  give  sufficierit  data  for  the  calculation  of  AB? 

4.  If  AC  =  500',  CD  =  320',  CE  =  180'  6",  calculate  AB. 

5.  Upon  what  principle  of  this  chapter  does  the  calculation  of  A  B 
depend? 

6.  How  does  this  method  differ  from  that  of  146,  Exercise  8? 

The  Distance  Between  Two  Points  on  Opposite  Sides  of  a 
Building  on  Pond. — The  distance  AB  is  required.  Set  a 
stake  at  any  point  C  from 
which  both  A  and  B  can 
be  seen;  measure  CA  and 
CB.  Mark  any  point  E  on 
CB  such  that  a  line  DE 
will  pass  the  building. 
Calculate  the  position  of 
point  D  on  CA  such  that 

DE  is  parallel  to  A  B,  and  set  a  stake  at  D.     Measure  DE 
and  calculate  A  B. 

EXERCISES 

7.  How  is  the  distance  CD  found,  such  that  DE  will  be  parallel 
to  AB? 

8.  How  is  the  distance  AB  calculated,  after  all  measurements  are 
completed? 

9.  Calculate  AB  if  CA  =  240',   CB  =  300',   CE  =  200',   DE  = 
238'. 

10.  Upon  what  principles  of  this  chapter  do  the  calculations  of  CD 
and  A  B  depend? 


MEASUREMENT    OF    SECTS,    AND    SIMILAR    TRIANGLES    141 

11.  How  does  this  method  differ  from  that  of  146,  Exercise  7? 

12.  Is  it  possible  to  modify  the  method,  by  setting  point  E  (a)  on 
CB  extended  through   B;   (6)   on   CB  extended  through   Cf 
Sketch  the  figures  in  these  cases. 

A  Perpendicular  to  a  Given  Line  at  a  Point  in  the  Line.— 
This  is  a  method  quickly  used  and 
often  sufficiently  accurate.     Mea-  pN^ 

sure   CD  =  40';   hold   a    100-foot  8j    ^?N 

tape  with  one  end  at  C  and   the    A"      c  "    ~~0X-^ — £ 
80-foot  mark  at  D;  take  hold  of 

the  30-foot  mark  and  hold  the  tape  so  that  both  parts  EC 
and  ED  are  taut.     Mark  point  E. 

EXERCISES 

13.  Upon  what  principle  of  this  chapter  does  the  method  depend? 

14.  What  is  the  largest  triangle  that  can  be  laid  out  in  this  way 
with  a  100-foot  tape,  with  exact  foot-lengths  for  its  sides? 

15.  Can  the  shorter  side  of  the  triangle  be  measured  on  CD? 

A  Perpendicular  to  a  Given  Line  from  a  Point  Without  the 
Line. — The  perpendicular  is  to  be  laid  out  from  C  which  is 
more  than  100  feet  from  line  AB. 
Place  any  stake  D  on  line  AB; 
measure  CD;  place  a  stake  E  on  AB 
at  about  the  foot  of  the  required 
perpendicular,  and  measure  DE; 
erect  a  perpendicular  EF;  place  a 
stake  at  G  on  both.EF  and  CD; 
measure  GC. 

EXERCISES 

16.  How  is  EH  calculated  such  that  CH    Q 
will  be  parallel  to  EG? 

17.  After    calculating    EH,    how    is    the    required    perpendicular 
marked  on  the  ground? 

18.  Let  CD  =  250.4',  DE  =  175.8',  GC  =  36.5';  calculate  EH. 

19.  May  the  point  E  be  placed  beyond  H  and  point  G  in  line  CD 
extended?     Draw  a  figure  illustrating  this  and  describe  the 
method  of  calculating  EH. 

20.  Upon  what  principle  of  this  chapter  does  the  method  depend? 


142 


PLANE    GEOMETRY 


Locating  a  Curved  Line. — A  straight  line  AE  is  staked  out 
near  the  curved  line;  and  divided  into  40-foot  intervals. 
Perpendiculars,  or  offsets,  CI,  DK,  etc.,  are  measured. 


EXERCISE 

21.  Plot  the  bank  of  a  creek  from  the  given  field  notes,  to  a  scale 
of  100  feet  to  the  inch.  The  points  lie  on  a  straight  line  at 
intervals  of  40  feet.  The  offsets  are  measured  to  the  center  of 
the  creek. 


Pts. 

Offsets 
at  40' 

Pts. 

Offsets 
at  40' 

Pts. 

Offsets 
at  40' 

Pts. 

Offsets 
at  40' 

0 

0 

3 

+  60 

6 

+60 

9 

-90 

1 

+  30 

4 

+100 

7 

+45 

10 

-35 

2 

+  50 

5 

+  95 

8 

-70 

CHAPTER  XI 
SIMILAR  TRIANGLES  AND  SIMILAR  POLYGONS 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

238.  The  First  Step  in  Classification. 

POSTULATES 

1.  Mutually  equiangular  triangles  are  similar. 

2.  The  homologous  sides  of  similar  triangles  are  pro- 
portional. 

3.  Two  triangles  are  similar  if  two  sides  of  one  triangle 
are  proportional  to  two  sides  of  the  other  and  if  the  included 
angles  are  equal. 

4.  Two  triangles  are  similar  if  all  the  sides  are  propor- 
tional. 

5.  Two  polygons  are  similar  if  they  are  mutually  equi- 
angular, and  also  if  their  homologous  sides  are  proportional; 
and  conversely. 

239.  The    Second    Step    in    Classification. — All    other 
principles  of  similar  triangles  and  similar  polygons  can  be 
analyzed  in  terms  of  the  postulates  of  238  and  other  previ- 
ously analyzed  principles. 

240.  Theorem  I. — A  proportion  can  be  changed  according 
to  certain  laws. 

(a)  Law  of  Products. — The  product  of  the  means  equals 
the  product  of  the  extremes. 

a      c 
Hypothesis. — Any  proportion, ,  =   ,• 

Conclusion. — ad  =  be. 

Help. — Clear  of  fractions.     What  axiom  is  used? 

Corollary. — If  the  product  of  two  numbers  equals  the 
product  of  two  other  numbers,  the  numbers  of  one  product  may 
be  made  the  means  and  those  of  the  other  product  the  extremes, 
of  a  proportion. 

143 


144  PLANE    GEOMETRY 

Hypothesis. — xy  =  st. 

Conclusion.— (1)  *  -  \    (2)  ~  =  *-,  (3)*  =  I  etc. 
by  i       y          jc        i 

Help. — (1)  Divide  by  sy;  etc. 

(6)  Law  of  Inversion. — The  ratios  of  a  proportion  may  be 

inverted. 

a      c 
Hypothesis. —  r  =  -v 

b      d 

Conclusion.—  -  =  -• 
u       c 

(c)  Law  of  Alternation. — The  means,  or  the  extremes,  of  a 
proportion  may  be  interchanged. 

Help. — Apply  law  (a) ;  divide  by  cd,  etc. 

(d)  Law   of   Composition. — The   sum   of  the   numerator 
and  denominator  of  each  ratio  may  be  made  the  numerators 

of  the  ratios. 

a      c 
Hypothesis. —  5  =  ^' 

a +  6      c +d 
Conclusion. —  — ^—       — ^— 

Help. — Add  1  to  each  ratio. 

(e)  Law  of  Division. 

a  —  b      c  —  d 
Conclusion. —  — ^ —  =  —^ — 

(/)  Law  of  Ratio  Series. 

ace 
Hypothesis. —  ^  =  ^  =  ->=  etc. 

a  +  c  +  eH a      c 

Conclusion.—  ^_  d  +  f  +  =  b  =d  = 

a          c 

Analysis.— (1)  ^  =  r;^  =  r;  etc. 

(2)  a  =  br;  c  =  dr;  etc. 

(3)  a  +  c  +  e  +  -        -  =br  +  dr  +  fr  +  • 

a  +  c  +  e 

(4)  r  =  t^+j^r. 

What  axioms  are  used? 


SIMILAR   TRIANGLES    AND    SIMILAR   POLYGONS  145 

EXERCISES 

3       4 

1.  Solve  these  proportions  for  the  unknown  term:  (1)  ^  =  -;   (2) 

?  _?    ,^*-4    m-5-31 
2  -  3,  W  g  -  3>  ^x  -  3 

Help.  —  Apply  law  (a). 

2.  Solve  for  the  unknown  term:  (1)  -  =  =;  (2)  ~  =  —  • 

X  /  £  ju 

Help.  —  Apply  law  (a). 

7       21 

3.  Change  the  proportion  5  =  ~=;   according   to    laws    (a)  to  (/). 

;J          Z  / 

Is  each  new  proportion  obtained  a  true  equation? 

4.  Change  the  ratio  series  ~  =  «  =  Tc  =  i~ci  according  to  law  (/). 

5.  Solve  x  :  re  +  1  =5:7  for  x,  and  check. 

6.  Show  that  the  proportion  T  =  -,  is  equivalent  to:  (1)  T  =  -;(2) 

q  c       ,         a  c  a+c        « 


c  +d       d 

241.  Theorem  II.  —  A  line  parallel  to  a  side  of  a  triangle 
divides  the  other  two  sides  proportionally. 

B 


Hypothesis.—  AABC  with  DE\\BC. 
„      ,     .          AD      AE 

Conclusion.- 


Helps.  —  Show  that  there  are  two  similar  triangles;  use 
Postulate  2,  238.     Draw  the  triangles  separately. 
DE      AD    AE 


AD  _AI3     AD  _  AE    AD      DB 
(2)  AE  ~  AC'    DB  ~  EC'   AE  ~  EC'  e 

rfAD      AE  AB       AC    AB  -  AD 

.-li-      =        ',   then 


AC-AE          AD  AE          AD  _  AE 

AE      '   AB-AD~  AC-AE'   DB~EC 

10 


146  PLANE    GEOMETRY 

242.  Theorem  III. — A  line  which  divides  two  sides  of  a 
triangle  proportionally  is  parallel  to  the  third  side. 


Hypothesis. — AABC  with  DE  drawn  so  that  -,-g  =  -j-~- 

Conclusion.— DE\\BC. 

Helps.— (1)  AADE  ~  AABC;  Postulates;  (2)  .'.  x  = 
y;  (3)  .'.  DE\\BC;  why?  Draw  the  triangles  separately. 

243.  Theorem  IV. — Two  triangles  similar  to  the  same 
triangle  are  similar  to  each  other. 


Hypothesis.— &DEF  and  GUI  ~  AABC. 

Conclusion.— ADEF  ~  AGHI. 

Helps.— LA  =  £D;  ZA  =  Z.G;  .'.  ZD=  Z.G;  etc. 

244.  The  Third  Step  in  Classification.— The  principles 
which  follow,  245  to  256,  are  probably  the  result  of  the 
application  of  deductive  reasoning,  and,  with  the  exception 
of  250  in  an  elementary  form,  were  not  discovered  experi- 
mentally. 

In  making  use  of  similar  triangles  in  analysis  it  is  ad- 
visable to  mark  the  mutually  equal  angles  of  the  two 
triangles  under  consideration,  x  and  x',  y  and  y',  z  and  z'. 
The  homologous  sides  are  then  readily  identified  as  being 
opposite  the  homologous  angles.  In  some  figures  it  is  also 
helpful  to  draw  the  triangles  under  consideration  apart 
from  the  general  figure,  and  to  letter  all  angles  and  sides  as 
they  are  lettered  in  the  general  figure.  The  use  of  colored 


SIMILAR   TRIANGLES   AND    SIMILAR    POLYGONS 


147 


crayons,  overlaying  homologous  sides  of  similar  triangles 
with  the  same  color,  will  be  found  a  great  help. 

Notice  that,  if  two  angles  of  one  triangle  are  equal,  each 
to  each,  to  two  angles  of  another  triangle,  the  triangles  are 
similar.  Why? 

245.  Theorem  V. — The  bisector  of  an  angle  of  a  triangle 
divides  the  opposite  side  into  segments  proportional  to  the 
adjacent  sides. 


State  the  hypothesis  and  the  conclusion. 

Helps.  —  (1)  How  are  the  auxilliary  lines  drawn?     (2) 

CA      CD 

-rjj,  =  jyF>',  why?     (3)    A  ABE  is  isosceles;   how  shown? 

C  A        CD 

(4)  .'.  AB  =  AE;  (5)  . 


Draw  the  elementary  figures  which  make  up  the  figure  of 
the  theorem. 

246.  Theorem  VI.  —  The  bisector  of  an  exterior  angle  of  a 
triangle  divides  the  opposite  side  externally  into  segments,  etc. 
(complete  the  theorem  after  analysis). 


CA       CD 
Helps.— (1}   Draw  BE;  (2)  -^  =  =^;   (3)    AABE    is 


isosceles;     (4)    AB  =  AE;    etc. 

Draw  the  elementary  figures  as  in  245. 


148 


PLANE    GEOMETRY 


247.  Theorem  VII. — The  altitude  drawn  to  the  hypotenuse 
of  a  right  triangle  divides  the  triangle  into  two  triangles  which 
are  similar  to  the  given  triangle  and  to  each  other. 


•c     D 


•c     0 


Helps. — The  three  triangles  to  be  considered  are  ABC, 
BDC  and  ADB. 

248.  Theorem  VIII. — The  altitude  drawn  to  the  hypotenuse 
of  a  right  triangle  is  a  mean  proportional  between  the  segments 
into  which  it  divides  the  hypotenuse. 

Helps. — Each  pair  of  triangles  of  247  affords  three  pro- 
portions. If  these  are  stated  in  full,  it  will  be  found  that  a 
proportion  with  a  repeated  mean  (or  extreme)  is  found  in 
each  set  of  three.  One  of  these  mean  proportionals  is  the 

a       c 
required  relation.     A  repeated  extreme,  T  =  -,  may  also  be 

regarded  as  a  mean  proportional,  since  an  equivalent  pro- 

b      a 

portion  is  -  =  — 
a       c 

249.  Theorem  IX. — Each  perpendicular  side  of  a  right 
triangle  is  a  mean  proportional  between  the  hypotenuse  and 
the  projection  of  the  side  upon  the  hypotenuse. 

Help. — The  required  relations  are  found  among  the 
proportions  suggested  in  248. 

Corollary. — The  square  of  a  perpendicular  side  of  a  right 
triangle  equals  the  product  of  the  hypotenuse  and  the  projection 
of  the  side  upon  the  hypotenuse. 

250.  Theorem  X. — The    sum  of  the  squares  of  the  two 


SIMILAR   TRIANGLES   AND    SIMILAR    POLYGONS 


149 


perpendicular  sides  of  a  right  triangle  equals  the  square  of  the 
hypotenuse. 


8 


Hypotenuse. — Right  AABC. 
Conclusion. — b2  —  a2  +  c2. 
Analysis. 


STATEMENT 

1.  a2  =  bn  and  c2  =  bm 

2.  a2  +  c2  =  bn  -f  6m 

3.  a2  +  c2  =  6(n  +  w) 

4.  a2  +  c2  =  6.6 

5.  .*.  a2  +  c2  =  62 
Corollaries.— (1)  a2  =  6 
(2)  a  = 


-  c 


REASON 

Theorem  IX,  Corollary 
Axiom 

Algebraic  Axiom 
Axiom 

Algebraic  Axiom 
;2  =  62  -  a2. 


c)(&-  c);  etc. 

The  relation  of  Corollary  2  is  convenient  in  numerical 
calculation. 

251.  Historical    Interest    of    Theorem    X.  (250).—  This 
relation  is  supposed  to  have  been  known  for  simple  number 
values  (at  least  for  the  numbers  3,  4,  5)  by  the  Egyptians 
as  early  as  2000  B.  C.,  and  also  by  the  ancient  Hindoos 
and  Chinese.     The  general  principle   was  discovered  by 
Pythagoras   (582-500  B.   C.),   whose  method  of  analysis 
was  probably  somewhat  like  the  method  here  presented. 
The  theorem  is  known  as  the  Pythagorean  Theorem.     An 
entirely  different  method  was  given  by  Euclid   (Chapter 
XVII).     Numbers  which  possess  the  relation  a2  +  62  =  c2 
are  called  Pythagorean  numbers;  many  such  series  exist 
in  higher  values. 

252.  Theorem  XI.  —  In  any  triangle,  the  square  of  a  side 
opposite  an  acute  angle  equals  the  sum  of  the  squares  of  the 
other  two  sides  minus  twice  the  product  of  one  of  those  sides 
by  the  projection  of  the  other  side  upon  it. 


150 


PLANE    GEOMETRY 


B 


FIG.  1.  FIG.  2. 

Hypothesis. — A  ABC  with  an  acute  /.A,  and  BD±AC. 

Conclusion. — a2  =  62  +  c2  —  26m. 

Analysis. 

STATEMENT 

1.  m  +  n  =  6,or  6  -f-  n  =  m 

2.  n   =  b   —  m  or  n  =  m  —  b 

3.  n2  =  52  _  26m  +  m2 


REASON 
Axiom  11 
Axiom  3 
Algebraic  axiom 


4.  n2  +  p2  =  62  -  26m  +  m2  +  p2  Axiom  2 

5.  a2  =  52  _  26m  +  c2  250  and  Axiom  13 
253.  Theorem  XII. — In  any  obtuse  triangle,  the  square 

of  the  side  opposite  the  obtuse  angle  equals,  etc.     (Complete  the 
theorem  after  analysis.) 

B 


Write  the  hypothesis  and  conclusion. 

Helps. — (1)   n  —  m  =  6;   (2)   .*.  n  =  b  +  m.     Complete 
as  in  252. 

254.  Theorem  XIII. — The  projection 
of  a  sect  can  be  calculated  when  the  angle 
which  the  sect  makes  with  the  line  of  pro- 
jection is  (a)  60°,  (6)  30°,  (c)  45°. 

Part  (a). 

Hypothesis. — A  sect  s  making  an 
angle  of  60°  with  the  line  of  projection 
AD,  and  its  projection  m. 


SIMILAR   TRIANGLES   AND    SIMILAR    POLYGONS  151 


Conclusion. — m  =  f  s. 
Analysis. 

STATEMENT 

1.  ZC  =  90°;  Z5  =  30° 

2.  m  =  |s  =  .5s 

Part  (6). 


REASON 
143,  Theorem  (8) 


B 


A  CO 

State  hypothesis  and  conclusion. 


3s2 


Helps.—  (1)    p  =  is;    (2)  m2  =  s2  -  p2;    (3)  m2  =  ^~  ; 


(4)  m  =  =  .866s. 

« 

Part  (c). 


C    D 


s\/2 


Helps.— (1)   m  =  p;  (2)  m2  +  p2  =  s2;  (3)  m  = 
=  .707s. 

255.  Theorem  XIV. — An  altitude  of  a  triangle  can  be 
expressed  in  terms  of  the  three  sides. 


AH -6 M 

« 1\ 


152  PLANE    GEOMETRY 

Analysis. 

STATEMENT 

1.  In  ABDA,  Fig.  1,  pb2  =  c2  -  m2. 

2.  In  AABC,  a2  =  62  +  c2  -  2bm 

A2      I      r2    _    «2 

3.  From  2,  m  =  D   ±^- 

4.  Substituting  from  3  in  1, 


pb 


5. 


=,.-(»- 
.(.+i 


26 

c2  - 


REASON 

What  theorem? 
What  theorem? 

Axioms 
Axiom 


26 


X 
ck  — 


/          62  +  c2  -  a2\      41  -  u    . 

f  c — sr—     —  J      Algebraic  axiom 


(a  -  6  +  c)(a  +  6  -  c) 
—  ^~-  — 


, 


A1     ,      . 
Algebraic  axiom 


7.  Let  a  +  6  +  c  =  2s 
then 

—  a  +  6  +  c  =  2s  —  2a  =  2(s  —  a)      Algebraic  axiom 
a  —  6  +  c  =  2s  —  26  =  2(s  —  6),     Algebraic  axiom 
etc. 

8.  Substituting  from  7  in  6 

2s-2(s  -  a)-2(s  -  6)-2(s  -  c) 
b   =  ~  Axiom 


.  _ 
9.  pb  =  T  \/s(s  —  a)(s  —  b)(s  —  c)          Algebraic  axioms 

Show  that  the  same  result  is  obtained  in  Fig.  2. 
Helps.—  (1)    pb2  =  c2  --  m2;    (2)    a2  =  62  +  c2  +  26m; 

a2-62-c2 
(3)  m  =  --  25  --  >'  etc- 

Write  the  values  of  pa  and  pc  by  interchanging  letters  in 
the  formula  for  pi. 

256.  Theorem  XV.  —  A  median  of  a  triangle  can  be  ex- 
pressed in  terms  of  the  three  sides. 

Analysis. 

1.  In  ABED,  pb2  =  mb*  -  t* 

2.  In  ABEC,  a2  =  pb2  +       +  t      =    b2  +       +  bt 


SIMILAR   TRIANGLES   AND    SIMILAR   POLYGONS  153 


3.  In  ABEA,  c2  =  pb2 

4.  Adding  2  and  3;  a2  +  c2  =  2p62  +      + 


5.  Substituting  value  of  pb2  from  1  in  4; 

62 
a2  +  c2  =  2(w62  -  t2)  +  -2  +  2*2  = 

2(a2  _|_  C2)  ._  52 

6.  Solving  5  for  m^\m^  •  ^ 

Show  that  the  same  result  is  obtained  if  Z.A  is  obtuse. 
Write  the  values  of  ma  and  mc  by  interchanging  letters  in 
the  formula  for  mb. 

257.  Theorem  XVI. — Two  similar  polygons  can  be  divided 
into  the  same  number  of  triangles  similar  each  to  each  and 
similarly  placed. 

B 


0  0 

Helps.— (I)  AABC~AA'B'C';  238,  Postulate  3;— (2)  .'. 
Z  1  =  Z2  and  Z3  =  Z4;  and  -p^r,  =  -^-,-g,  =  ^7^>; — (3)  /5 
=  Z6,  why?  and  /.  &ACD~AAC'D';  etc. 


154  PLANE    GEOMETRY 

258.  Theorem  XVII. — Two  polygons  that  are  composed  of 
the  same  number  of  triangles  similar  each  to  each  and  simi- 
larly placed,  are  similar. 


Helps.— (I)  ZA  =  ZA';  Zl  =  Z2;  Z3  =  Z4;—  (2)  Z5  =  Z6; 
Z7  =  Z8;  etc., — (3)  .'.  polygons  are  mutually  equiangular. 

...    AB        BE        BC        EC  . 

~(4)  A7B'  =  WE'  =  PC7  =  WC'  =  etc--®  '  •  Polygons 
have  their  homologous  sides  proportional.- — (6)  .'.  the  poly- 
gons are  similar;  238,  Postulate  5. 

259.  The  Fourth  Step  in  Classification. — The  principles 
of]  245  to  256  cannot  be  considered  as  established  beyond 
all  doubt  until  they  have  been  tested  by  actual  measure- 
ments. This  we  will  now  proceed  to  do. 

EXERCISES 

1.  Draw  a  triangle  (full  size)  whose  sides  are  4,  5.6  and  6.2  inches. 
Draw  the  bisector  of  each  interior  angle.     Calculate  by  245 
the  segments  into  which  each  bisector  divides  the  opposite  side. 
Measure  these  segments  and  note  the  agreement  of  the  meas- 
urements with  the  calculated  values. 

Thus  to  calculate  the  segments  of  the  largest  side,  calling  the  seg- 
ments x  and  6.2  —  x,  we  have  7  =    '  -  „ — ,  whence  x  =  2.58", 

and  6.2  —  re  =  3.62".     Calculate  in  the  same  way  the  segments 
of  the  other  sides. 

2.  Draw  the  bisectors  of  each  pair  of  exterior  angles  of  the  triangle 
of  Exercise  1,  and  prolong  each  bisector  until  it  meets  the  oppo- 
site side  prolonged.     Calculate  by  246  the  segments  into  which 
each  bisector  divides  the  opposite  side  externally.     Measure 
these  segments  (from  the  external  point  of  division  to  each  end 
of  the  side),  and  note  the  agreement  with  the  calculated  values. 

Thus  to  calculate  the  segment  of  the  largest  side,  calling  the 
segments  x  (to  the  nearer  end  of  the  side)  and  6.2  +  x  (to  the 

farther  end  of  the  side),  we  have  ^  =     '  ,  fi — »  whence   derive 


SIMILAR   TRIANGLES   AND   SIMILAR   POLYGONS  155 

values  of  x  and  6.2  +  x.     Calculate  in  the  same  way  the  seg- 
ments of  the  other  sides. 

3.  Draw  a  right  triangle  (full  size)  whose  sides  are  3,  4  and  5  inches. 
Draw  the  altitude  to  the  hypotenuse.     Calculate  by  249  the 
projection  of  each  perpendicular  side  upon  the  hypotenuse,  i.e., 
the  segments  into  which  the  altitude  divides  the  hypotenuse. 
Measure  these  projections  (or  segments)  and  note  the  agree- 
ment with  the  calculated  values. 

Thus  to  calculate  the  smaller  segment  x,  of  the  hypotenuse,  we 

have  I  =  „  whence  x  =  1.8".     Calculate  in  the  same  way  the 

other  segment  y. 

4.  In  the  triangle  of  Exercise  3,  calculate  by  248  the  length  of  the 
altitude  drawn  to  the  hypotenuse.     Measure  the  altitude  and 
compare  with  the  calculated  value. 

5.  Draw  a  sect  3  inches  long  and  through  one  end  of  it  a  line  mak- 
ing with  the  sect  an  angle  of  60°.     Project  the  sect  on  the 
second  line.     Calculate  by  254(a)  the  value  of  the  projection. 
Measure  the  projection  and  compare  with  the  calculated  value. 

6.  Draw  a  sect  of  4  inches,  and  a  line  making  an  angle  of  30°  with 
it.     Project  the  sect  on  the  other  line.     Calculate  by  264(6) 
the  value  of  the  projection.     Measure  and  compare  results. 

7.  Draw  any  sect  and  a  line  making  an  angle  of  45°  with  it.     Pro- 
ceed as  in  Exercises  5  and  6. 

8.  Draw  a  triangle  (full  size)  of  which  two  sides  are  5  and  3  inches 
and  the  included  angle  60°.     Calculate  the  third  side  by  252. 

Thus:  the  projection  of  the  3"  side  on  the  5" side  is  (by  254(a)) 
1.6". 

Then  a2  =  32  +  52  -  2(5)  (1.5).  Complete  the  calculation. 
Measure  the  third  side  and  compare  with  the  calculated  value. 

9.  Draw  two  triangles  in  each  of  which  two  sides  are  5  and  3  inches 
respectively;  in  one  triangle  the  included  angle  being  30°,  and 
in  the  other  45°.     Calculate  the  projections  of  the  3"  side  on 
the  5"  side  by  254(6)  and  254 (c),  and  complete  calculations  for 
the  third  side  of  each  triangle  as  in  Exercise  8. 

10.  Draw  a  triangle  of  which  two  sides  are  5  and  3  inches  and  the 
included  angle  120°.     Calculate  the  third  side  by  253. 

Thus:  the  projection  of  the  3"  side  on  the  5"  side  is  1.5". 
Then  a2  =  32  +  52  +  2(5)  (1.5);  complete  the  calculation. 
Measure  the  third  side  and  compare  with  the  calculated  value. 

11.  Draw  triangles  with  the  sides  as  in  Exercise  9,  the  included 
angle  in  one  triangle  being  150°  and  in  the  other  135°.     Calcu- 
late the  projections  of  the  3"  side  on  the  5"  side  as  hi  Exercise  9, 
and  the  values  of  the  third  side  of  each  triangle  by  253.     Meas- 
ure and  compare  values. 


156  PLANE    GEOMETRY 

12.  Draw  a  triangle  (full  size)  whose  sides  are  4,  5,  and  6  inches. 
Draw  the  three  altitudes.     Calculate  the  lengths  of  the  alti- 
tudes by  265. 

Thus,  to  calculate  the  altitude  to  the  longest  side: 

4+5+6 

•          —  Q  —          I     .O 

s  -  4  =  3.5,  s  -  5  =  2.5,  s  -  6  =  1.5 

P  =  I  V(7.5)(3.5)(2.5)(L5) 

Complete  the  calculation.  Measure  the  altitude  and  note  the 
agreement  with  the  calculated  value.  Calculate  and  measure 
the  other  altitudes  in  the  same  way. 

13.  Draw  the  medians  in  the  triangle  of  Exercise  12.     Calculate 
their  lengths  by  266.     Measure  these  lengths  in  the  triangle 
and  compare  with  the  calculated  values. 

14.  In  every  case  in  the  preceding  exercises  do  the  calculated  values 
show  a  sufficiently  close  agreement  with  the  deduced  principles 
as  to  confirm  the  accuracy  of  these  principles? 

260.     ADDITIONAL  THEOREMS 

(1)  Two  isosceles  triangles  with  equal  vertex  angles  are 
similar. 

(2)  All  equilateral  triangles  are  similar. 

(3)  Homologous  altitudes  of  similar  triangles  are  propor- 
tional to  any  two  homologous  sides. 

Help. — Show  that  the  triangles  containing  the  altitudes 
as  homologous  sides  are  mutully  equiangular. 

(4)  Homologous  medians  of  similar  triangles  are  propor- 
tional to  any  two  homologous  sides. 

Help. — Use  238,  Postulate  3,  to  show  that  the  triangles 
under  consideration  are  similar. 

(5)  Homologous   angle-bisectors   of  similar   triangles   are 
proportional  to  any  two  homologous  sides. 

(6)  In  a  triangle  ABC,  altitudes  AD,  BE  and  CF  are  drawn 
Show  that   AADC  is  similar  to   ABEC,  and    AADB  is 
similar  to  ACFB,  and  that  there  is  another  pair  of  similar 
triangles  formed. 

(7)  Two  right  triangles  are  similar  if  they  have  a  pair  of 
equal  acute  angles. 

(8)  Two  isosceles  triangles  are  similar  if  they  have  a  base 
angle  of  each  respectively  equal. 


SIMILAR    TRIANGLES    AND    SIMILAR    POLYGONS 


157 


(9)  The  projections  of  the  segments  of  a  sect  are  propor- 
tional to  the  segments  (232). 

(10)  A  system  of  parallel  lines  divides  any  two  transversals 
proportionally. 

(11)  The  diagonals  of  a  trapezoid  divide  each  other  propor- 
tionally. 

(12)  State  another  theorem  relating  to  the  proportion- 
ality of  the  segments  into  which  the  diagonals  of  a  trapezoid 
divide  each  other. 

(13)  A    pencil   of  three   concurrent  lines 
divides  two  parallel  lines  which  are  cut  by  all 
the  lines  of  the  pencil,  proportionally. 

Helps. — Use  two  pairs  of  similar  tri- 
angles. The  two  proportions  thus  occur- 
ring have  a  common  ratio. 

(14)  The  diagonal  of  a  square  equals  the  side  multiplied  by 
\/2;  or  d  =  s\/2. 

(15)  The  squares  of  the  perpendicular  sides  of  a  right 
triangle  are  proportional  to  their  projections  on  the  hypotenuse. 

Help. — Apply  249,  corollary. 

(16)  The  altitude  of  an  equilateral  triangle  may  be  expressed 

s\/3 
in  terms  of  the  side;  or  p  -   — - — . 

(17)  The  side  of  a  square  may  be  expressed  in  terms  of  the 
diagonal. 

(18)  The  side  of  an  equilateral  triangle  may  be  expressed  in 

terms  of  the  altitude;  or  s   =        ^    . 

o 

(19)  Find  a  method  of  drawing 
a  square  HGKL  inscribed  in  the 
triangle  ABC.      Prove    that    the 
method  found  is  correct. 

Helps.— (I)    AF  =  FE;—(2) 

°^~  -  —  -     m    ?P  -  ?.  (4\   • 
FE       BF~(6)    AF  ~ 

GL  _HG 
FE  ~AF;  etc' 


158 


PLANE    GEOMETBY 


(20)  A  relation  exists  between  a  perpendicular  on  one  side 
of  a  diameter  of  a  circle  and  the  segments  into 

which  it  divides  the  diameter. 

Helps. — Draw  DA  and  DB;  see  143,  Theorem 
(7).  Then  apply  one  of  the  preceding  theorems 
of  this  chapter  relating  to  right  triangles. 

(21)  A  relation  exists  between  a  chord  of  a  circle,  the  diam- 
eter drawn  through  the  end  of  the  chord,  and  the  projection  of 
the  chord  on  that  diameter. 

(22)  A  relation  exists  between  the  product  of  the  perpendicu- 
lar sides  of  a  right  triangle,  the  hypotenuse,  and  the  altitude  to 
the  hypotenuse. 

(23)  Any  angle  of  a  triangle  may  be  evaluated,  whether  acute, 
right  or  obtuse,  by  the  relation  of  the  squares  of  the  three  sides. 

(24)  Two  triangles  whose  sides  are  parallel,  each  to  each, 
are  similar. 

Helps. — From  81,   we  may  have;   (a)   x  -\-  xf  =  180°, 
y  +  y'  =  180°,  z  +  z'  =  180°;  or  (b)  x  +  x'  =  180°,  y  + 


y'  =  180°,   z  = 


or   (c)   x  +  X*  =  180°,  y  =  y 


=  z 


or  (d)  x  =  x'  ,  y  =  yf,  z  —  zr.     Show  that  (d)  is  the  only 
possible  numerical  relation  of  the  angles  of  the  two  triangles. 

(25)  A  relation  exists  between  two  triangles  whose  sides  are 
perpendicular,  each  to  each. 

(26)  The  segments  into  which  the  bisector  of  angle  A  of 

A  „  ~    ,  .  .  ,        .  ,  ab         _    ac 

A  ABC  divides  side  a,  are  =—  -  and  r—  ;  — 

b  +  c         b  +  c 

x         b 


Help. — If  x  is  a  segment  of  a, 


(27)  Point  D  lies  within 
LA.BC.     Find  a  method  of 
drawing  a  sect  through  point 
Determinated  by  the  sides  of 
the   angle   and  bisected  by 
point   D.     Prove    that  the 
method  found  is  correct. 

(28)  In    the    figure     of 
Theorem  (27) ,  find  a  method 


SIMILAR   TRIANGLES   AND    SIMILAR   POLYGONS  159 

of  drawing  a  sect  through  point  D  so  that  it  is  divided  by  point 
D  in  any  given  ratio.    Prove  that  the  method  found  is  correct. 

(29)  The  product  of  a  side  of  a  triangle  multiplied  by  the 
altitude  drawn  to  that  side,  equals  the  product  of  any  other 
side  of  the  triangle  multiplied  by  the  altitude  drawn  to  it. 

(30)  Two  polygons  similar  to  the  same  polygon  are  similar 
to  each  other. 

Help. — It  is  necessary  to  prove  both  that  the  polygons 
are  mutually  equiangular  and  that  their  homologous  sides 
are  proportional. 

(31)  Homologous  diagonals  of  similar  polygons  are  pro- 
portional to  any  two  homologous  sides. 

(32)  The  perimeters  of  two  similar  polygons  are  propor- 
tional to  any  two  homologous  sides. 

Help.—(l)  The  sides  -,  =  ^  =  -t  =  etc.;  (2)  apply  240 
a       o       c 

co- 

(33)  A  sect  is  divided  harmonically  into  segments  of  which 
x  is  one  internal  segment  and  y  is  the  corresponding  external 

segment,  when  y  =     _  „  • 

(34)  The  bisectors  of  adjacent  interior  and  exterior  angles  of 
a  triangle  divide  the  opposite  side  harmonically. 

(35)  The  sum  of  the  squares  of  the  sides  of  a  parallelogram 
equals  the  sum  of  the  squares  of  the  diagonals. 

Help. — Express  the  value  of  each  diagonal  from  252  and 
253,  and  add. 

(36)  Four   times  the  sum  of  the 
squares    of   the    medians  drawn   to 
the  perpendicular  sides  of  a  right 
triangle,  equals  five  times  the  square 

of  the  hypotenuse.  & 

Helps. — (1)  Express  x  and  y  from  256 ;  (2)  square  these 
values  and  multiply  out  the  coefficients  2  and  J ;  (3)  add 
and  reduce. 

(37)  The  difference  of  the  squares  of  two  sides  of  any  triangle 
equals  the  difference  of  the  squares  of  the  projections  of  these 
sides  upon  the  third  side. 


160  PLANE    GEOMETRY 

Help. — Draw  an  altitude,   thus  forming  two  right  tri- 
angles. 

261.  REVIEW  EXERCISES 

1.  The  perpendicular  sides  of  a  right  triangle  are  12  and  16.     Cal- 
culate the  hypotenuse. 

2.  Calculate  the  diagonal  of  a  square  whose  side  is  10. 

3.  Calculate  the  side  of  a  square  whose  diagonal  is  12. 

4.  Two  sides  and  the  included  angle  of  a  triangle  are  10,  15,  135°. 
Calculate  the  third  side. 

5.  Two  sides  and  the  included  angle  of  a  triangle  are  20,  45,  60°. 
Calculate  the  third  side. 

6.  The  perpendicular  sides  of  a  right  triangle  are  6  and  10.     Calcu- 
late their  projections  on  the  hypotenuse. 

7.  To  check  the  layout  of  a  tennis  court,  36  feet  by  78  feet,  calcu- 
late and  measure  the  diagonals. 

8.  A  derrick  mast  50  feet  high  is  guyed  by  a  rope  running  from 
the  top  of  the  mast  to  a  tree  trunk  100  feet  distant.     Allow 
5  feet  for  fastening,  and  calculate  the  length  of  rope  required. 

9.  Calculate  the  resultant  of  two  forces  of  75  Ib.  and  32  lb.,  acting 
at  an  angle  of  90°. 

10.  The  resultant  of  two  forces  acting  at  an  angle  of  90°  is  12  lb.; 
one  of  the  forces  is  3  lb.     Calculate  the  other  force. 

11.  Calculate  the  resultant  of  two  forces  of  10  lb.  and  15  lb.  acting 
at  an  angle  of  45°;  at  an  angle  of  135°. 

12.  Calculate  the  resultant  of  three  forces  of  10,   12  and  15  lb. 
respectively,  the  first  two  forces  acting  at  a  right  angle  to  each 
other,  and  the  third  force  acting  at  a  right  angle  to  the  resultant 
of  these  two. 

13.  Show  that  the  values  a2  +  62,  a2  —  62,  2ab  may  represent  the 
sides  of  a  right  triangle.     Where  is  the  right  angle? 

14.  Pythagoras  stated  that  the  values  2n  +  1,  2n(n  +  1),  2n2  -f 
2n  +  1  are  the  sides  of  a  right  triangle.     Prove  that  this  is  so. 

15.  Let  n  =  1  in  the  general  values  of  Exercise  14;  show  that  the 
sides  of  the  triangle  are  in  this  case,  3,  4,  and  5.     Let  n  =  2, 
and  find  the  sides  of  the  right  triangle  from  the  formulas  of 
Exercise  14.     Similarly,  let  n  =  3,  4,  5,  etc. 

16.  One  of  the  perpendicular  sides  of  a  right  triangle  is  2  more  than 
the  other;  the  hypotenuse  is  10.     Calculate  the  sides.     Solve 
by  forming  an  algebraic  equation.     Check. 

17.  One  of  the  perpendicular  sides  of  a  right  triangle  is  20  inches 
and  the  hypotenuse  is  10  more  than  the  other  side.     Calculate 
side  and  hypotenuse. 


SIMILAR   TRIANGLES   AND    SIMILAR   POLYGONS  161 

18.  An  ancient  Chinese  problem  which  appears  in  many  texts:  In 
the  middle  of  a  circular  well  10  feet  in  diameter,  grows  a  reed 
which  projects  1  foot  above  the  surface.     When  the  reed  is 
bent  over,  remaining  straight  from  its  root,  the  top  just  reaches 
the  edge  of  the  well.     How  deep  is  the  well? 

19.  A  builder  plans  a  barn  with  the  ridge  of 
the  roof  10  feet  above  the  loft  floor,  and 
the  floor  24  feet  wide.      How  long  must 
he    order    the  rafters,   allowing    2    feet 
overhang  at  the  eaves  and  6  inches  for 
trimming  the  ends? 

20.  Two  chords  of  a  circle,  8  and  12  inches  respectively,  are  drawn 
from  the  same  point  in  a  circumference  to  the  extremities  of  a 
diameter.     Calculate  the  diameter. 

21.  A  perpendicular  to  the  diameter  of  a  circle  is  10  inches  and  one 
of  the  segments  into  which  it  divides  the  diameter  is  5  inches. 
Calculate  the  diameter. 

22.  The  altitude  to  the  hypotenuse  of  a  right  triangle  is  3  inches 
and  the  projection  of  one  of  the  perpendicular  sides  upon  the 
hypotenuse  is  8  in.     Calculate  the  three  sides  of  the  triangle. 

23.  The  rise  CB  in  100  feet  of  slope  AB 
is   17  feet.      Calculate  the  horizontal 
distance  in  100  ft.  of  slope.  __^ 

24.  A  40-foot  ladder  is  placed  against  a 

building  with  its  foot  10  feet  from  the  wall.     How  high  does 
it  reach  on  the  building? 

25.  Can  a  rod  40  inches  long  be  placed  within  a  trunk  24  inches  by 
30  inches,  by  16  inches  high;  from  an  upper  corner  to  a  lower 
opposite  corner? 

26.  The  sides  of  a  triangle  are  10,  25,  30.     Calculate  the  segments 
into  which  each  angle-bisector  divides  the  opposite  side. 

27.  Calculate  the  position  of  the  point  where  the  bisector  of  the 
exterior  angle  opposite  the  30  side  of  Exercise  26  intersects  this 
side  produced. 

28.  Calculate  the  length  of  the  three  altitudes  of  the  triangle  of 
Exercise  26. 

29.  Calculate  the  lengths  of  the  three  medians  of  the  triangle  of 
Exercise  26. 

30.  Two  men  are  walking  on  opposite  sides  of  a  street.     A  tree 
near  the  curb  is  between  them.     The  man  on  the  farther  side 
walks  uniformly;  must  the  man  on  the  near  side  walk  uniformly 
or  otherwise,  in  order  to  keep  the  tree  always  between  them? 

31.  Show  how  the  methods  of  these  figures  can  be  used  to  divide  a 
given  sect  4J3.  into  any  number  of  equal  parts.     Explain  the 
11 


162 


PLANE    GEOMETRY 


principles  on  which  the  result  depends.     Draw  similar  figures 
showing  the  division  of  the  sect  into  5  equal  parts. 

.5 


Line  DE  is  parallel  to 
A 


32.  Construct  a  right  triangle  when  the  sum  a,  of  the  two  perpen- 
dicular sides,  and  an  acute  angle  a 

or,  are  given.  H —  — i— 

33.  Construct  a  right  triangle  when 
the  perimeter   and   a  similar  tri- 
angle are  given. 

34.  Calculate  the  value  of  \/352  —  242 
by  the  method  of  250,  Corollary  2. 

35.  In   AABC,  AB  =  10",  AC  =  15". 
BC  and  AE  is  3"  more  than 

AD.  Find  the  lengths  AD  and 

AE.  Check  the  result. 

36.  In  the  triangle  of  Exercise  35, 
AE    is    10"   more   than   AD. 
Calculate  AD  and  AE  and  check  result. 

37.  Test  the  principle  of  260,  Theorem  (13),  by  accurate  measure- 
ment. 

38.  Test  the  principles  of  260,  Theorem  (14),  by  measurement. 

39.  Test  the  principle  of  260,  Theorem  (15),  by  measurement. 

40.  Test  the  principle  of  260,  Theorem  (16),  by  measurement. 

41.  Calculate  the  diagonal  of  a  square  if  the  side  =  50'. 

42.  Calculate  the  altitude  of  an  equilateral  triangle  if  a  side  is  120'. 

43.  Calculate  the  side  of  a  square  if  the  diagonal  is  24". 

44.  Calculate  the  side  of  an  equilateral  triangle  if  the  altitude  is  24". 

45.  Test  the  principle  of  260,  Theorem  (20),  by  accurate  measure- 
ment. 

46.  The  diameter  of  a  circle  is  24".     Calculate  the  length  of  the 
perpendicular  on  one  side  of  the  diameter  erected  at  a  point  6" 
from  one  end. 

47.  The  sides  of  a  triangle  are  10",  12",  16".     Calculate  the  seg- 
ments into  which  the  bisector  of  the  largest  angle  divides  the 
opposite  side,  using  the  formulas  of  260,  Theorem  (26). 

48.  Determine  which  angles  of  the  following  triangles  are  acute, 
right  or  obtuse.    .  (1)  a  =  5,  6  =  12,  c  =  13;  (2)  a  =  15,  b  =  10, 
c  =  18;  (3)  a  =  68,  b  =  51,  c  =  100;  (4)  a  =  7,  6  =  25,  c  =  24. 


SIMILAR   TRIANGLES   AND    SIMILAR   POLYGONS 


163 


51. 


52. 


49.  The  sides  of  a  right  triangle  are  3,  4,  5.     Calculate  (a)  the  alti- 
tude to  the  hypotenuse;  (6)  the  segments  into  which  the  altitude 
divides  the  hypotenuse;  (c)  the  length  of  the  median  to  the 
hypotenuse;  (d)  the  length  of  the  bisector  of  the  right  angle; 
(e)  the  segments  into  which  this  angle-bisector  divides  the 
hypotenuse. 

50.  To  what  simpler  formula  do  the  formulas  of  Theorems  XI  and 
XII  of  similar  triangles  reduce  if  A  is  a  right  angle?     What 
does  the  projection  of  c  on  6  then  become? 

Calculate  the  median  to  the  base  of  an  isosceles  triangle  whose 
sides  are  4",  4",  3";  (a)  using  the  formula  of  250  ;  (6)  using  the 
formula  of  255  ;  (c)  using  the  formula  of  256. 
A  carpenter  lays  out  a  perpendicular  by  measuring  a  string 
24  feet  long,  with  knots  at  8  feet  and  14  feet  from  one  end. 
The  ends  are  tied  together  and  the  string  is  stretched  around 
three  nails  so  that  the  knots  are  at  the  nails.  Make  a  sketch 
and  explain  the  principle  upon  which  the  method  depends. 

53.  A  man  5  feet  6  inches  tall  stands  8  feet  from  a  lamp-post  and 
casts  a  shadow  on  the  ground  10  feet  long.     How  high  is  the 
lamp-post? 

54.  Calculate  the  altitudes  of  a  triangle,  of  which  the  sides  are  102, 
104  and  106  feet. 

55.  What  is  the  diameter  required,  of  a  bar  of  round  tool  steel  in 
order  that  a  square  bar  f  inch  on  a  side  may  be  cut  from  it? 

56.  What  is  the  largest  square  bar  that  can  be  cut  from  a  round 
bar  of  steel  1$  inches  in  diameter? 

57.  Calculate  the  chord  of  a  circle  whose  radius  is  12",  which  sub- 
tends an  arc  of  90°. 

58.  Calculate  the  length  of  a  tangent  drawn  to  the  circle  of  Exercise 
57,  from  a  point  which  is  30"  from  the  center  of  the  circle. 

59.  Calculate  the  length  of  the  shortest  chord  which  can  be  drawn 
in  the  circle  of  Exercise  57,  through  a  point  10"  from  the  center 
of  the  circle. 

The  radii  of  two  concentric  circles  are  10"  and  12".     Calculate 

the  length  of  a  chord  of  the  larger  circle  which  is  tangent  to  the 

smaller. 

Show  that  the  side  of  a  regular  octagon  con- 

structed  by   cutting    the    corners    from    a 

square  is  s'  =  s(\/  2  —  1). 


60. 


61 


.—  s  =s'  +  2x  =  V  + 


;  solve  for  s'. 


62.  If  an  octagon  of  side  s'  is  required,  show 
that  it  may  be  drawn  by  first  drawing  a 
square  of  side  s,  such  that  s  =  s'(\/2  +  1). 


164 


PLANE    GEOMETRY 


63.  Find  the  side  of  a  regular  octagon  formed  by  cutting  the  corners 
from  a  square  whose  side  is  20. 

64.  What  is  the  side  of  a  square  which  must  be  first  drawn,  in  order 
that  the  regular  octagon  formed  by  cutting  off  the  corners  shall 
have  sides  equal  to  20? 

APPLICATIONS 

262.  Drawing  Maps  and  Designs  to  Scale. — All  reduc- 
tions and  enlargements  of  figures  depend  upon  the  following 
general  property  of  similar  figures:  If  all  angular  measure- 
ments of  two  figures  are  equal,  all  linear  measurements  are 
proportional;  and  conversely.  Enlargements  of  stage 
scenery,  wall  decorations,  architectural  designs,  etc., 
are  made  as  illustrated  in  the  following  exercises. 


EXERCISES 

1.  Measure  the  bearing  of  A  from  B  on  the  map  of  Fig.  1.  Measure 
the  distance  AB.  Measure  the  approximate  coast  line  of  the 
island. 


SCALE    OF   MILES 
05  10  15 


FIG.  1- 


FIG.  2. 


FIG.  3. 


2.  Enlarge  Fig.  2  on  a  scale  of  1  to  3.     Draw  a  similar  rectangle, 
divided  into  the  same  number  of  squares,  and  follow  proportion- 
ally located  points  in  the  enlarged  figure. 

3.  Enlarge  the  design  of  Fig.  3  on  a  scale  of  1  to  2  5. 


SIMILAR   TRIANGLES   AND    SIMILAR    POLYGONS 


165 


The  pantograph  is  an  instrument  for  enlarging  or  reducing 
drawings.     A  simplified  form  of  the  instrument  is  shown. 


EXERCISES 

4.  Sketch  the  pantograph  adjusted  so  as  to  enlarge  in  a  ratio  of 
1  to  5. 

5.  How  is  the  instrument  used  if  a  design  is  to  he  reduced  to  one- 
fifth  the  original  size? 


Proportional  Dividers. — This  instrument  is  used  by 
draughtsmen  to  obtain  a  sect  having  a  given  ratio  to  a 
given  sect.  It  can  thus  be  used  to  transfer  a  drawing,  one 
measurement  after  another,  to  a  different  scale. 

EXERCISE 

6.  Explain  the  operation  of  the  instrument. 

263.  THE  PLANE  TABLE. — This  is  an  instrument  used  in 
surveying  to  locate  points  within  or  adjacent  to  a  plot  of 
ground,  as  corners  of  buildings,  trees,  fence  corners,  etc., 


166 


PLANE    GEOMETRY 


upon  a  map.  It  consists  of  a  drawing  board  mounted  on  a 
tripod.  One  line,  as  AB,  of  a  plot  is  measured  and  drawn  to 
scale  on  a  sheet  of  paper  which  is  fastened  to  the  Plane 
Table.  The  table  is  set  up  in  the  field  over  each  end  in  turn 
of  the  line  AB.  The  lines  are  drawn  on  the  paper  in  the 
directions  of  the  points  to  be  located.  The  angular  values 
are  thus  preserved  on  the  paper,  and  the  resulting  figure  is 
similar  to  the  true  figure  lying  on  the  ground. 

Objects  in  a  lake  or  bay  may  be  located  with  the  Plane 
Table. 


>c 


264.  THE  GEOMETRIC  SOLUTION  OF  NUMERICAL  PROB- 
LEMS.— These  methods  are  chiefly  interesting  because  of 
their  antiquity,  since  they  were  used  by  Greek  mathemati- 
cians many  centuries  before  arithmetic  was  sufficiently  per- 
fected and  before  algebra  had  been  invented. 

Any  unit  may  be  used  in  measuring  the  sects,  the  same 
unit  being  used  throughout  a  problem. 

A  Fourth  Proportional  of  Three  Given  Numbers. — Let  the 
three  given  numbers  be  represented  by  sects  a,  6  and  c. 
Draw  two  intersecting  lines  MN  and  MP;  lay  off  the  sects 
a,  6  and  c,  as  shown;  draw  QR  and  ST  parallel  to  QR. 

Then  x  is  the  fourth  proportional  required,  since  -r  =  -• 

ox 

In  order  to  find  the  numerical  value  of  the  result,  measure  x 
with  the  same  unit  used  in  drawing  the  given  sects,  a,  6,  c. 


SIMILAR   TRIANGLES    AND    SIMILAR   POLYGONS 


167 


b  K 


C   H 


The    Product    of    Two    Numbers. — Let    db 

,  ,  1       b 

ab  =  a?-l;and-  =  -- 


a      x 


The  Quotient  of  Two  Numbers. — Let  T  =  x;  then  r  =  ~' 
Draw  the  figure  as  in  the  preceding  problems. 


168  PLANE    GEOMETRY 

A  Number  Divided  into  Parts  Proportional  to  Two  Given 
Numbers. — Let  it  be  required  to  divide  line  a  into  parts  pro- 
portional to  m  and  n. 


n  i- 


EXERCISES 

Solve  these  problems   by  the  geometric  method  and  check  the 
result  arithmetically: 

1.  Find  the  fourth  proportional  of  2,  5,  8. 

2.  Find  the  third  proportional  of  2  and  3. 

3.  Find  the  product  of  2.5  and  1.8. 

4.  Find  the  square  of  2.6. 

5.  Find  the  quotient  of  4.3  by  1.7. 

6.  Divide  7  into  parts  proportional  to  2  and  3. 

The    Mean   Proportional    Between    Two    Numbers. — Let 
a  and  b  be  the  given  numbers.     It  is  required  to  find  the 

value  of  x  in  the  proportion      =  v. 


Show  that  this  value  is  determined  in  the  constructed 
figure  (260,  Theorem  (20)). 

The,  Square  Root  of  a  Number. — Let  x  =  \/a;  then  x2  — 

a;  and  x.x  =  a.l;  -  =     .     Construct  the  figure  as  in  the 
x       a 

preceding  problem,  assuming  a  unit  of  measurement  u. 


SIMILAR   TRIANGLES   AND    SIMILAR    POLYGONS  169 

EXERCISES 

7.  Find  the  mean  proportional  of  2  and  8. 

8.  Find  the  square  root  of  4. 

9.  Find  the  mean  proportional  of  1.5  and  2.4. 

10.  Find  the  square  root  of  3. 

11.  Solve  geometrically,  x  =  \/2.7  X  4.5. 

A  Table  of  Square  Roots. — The  accompanying  figure  may 
be  regarded  as  a  geometric  table  of  square  roots.  The 
diagonal  lines  give  the  square  roots  of  consecutive  integers. 


EXERCISES 

12.  Explain  why  x  =  \/2;  y  =  \/3;  etc. 

13.  Construct  the  figure  as  far  as  the  square  root  of  6.     Make  a 
numerical  table  and  compare  with  the  known  arithmetical 
values. 

265.  FIRST  PRINCIPLES  OF  TRIGONOMETRY. — This  great 
science  has  been  referred  to  in  several  places.  It  has  been 
developed  from  the  principles  of  similar  triangles.  If  a 
sufficient  number  of  lines  and  angles  of  a  figure  are  known  to 
permit  the  definite  construction  of  the  figure,  then  the 
remaining,  or  unknown  parts,  may  be  determined  by 
measuring  the  completed  figure.  The  object  of  trigonometry 


170 


PLANE    GEOMETRY 


is  to  furnish  methods  of  calculating  the  unknown  parts  with 
any  desired  degree  of  accuracy. 

The  Functions  of  an  Acute  Angle. — Draw  any  acute  angle 
x;  draw  a  line  0  perpendicular  to  side  A.  In  the  triangle 
HO  A,  the  ratio  of  any  two  sides  is  called  a  function  of 
angle  x. 


Four  of  the  six  functions  are  called: 

side  opposite 
hypotenuse 


sine  x 


0 
H 


side  adjacent      A 

cosine  x  =  -r—  ~  =  7? 

hypotenuse       H 

0 
A 


_  side  opposite 
side  adjacent 

side  adjacent      A 

cotangent  x  =    . ,  — r—  =  7: 

side  opposite      0 


EXERCISES 

1.  Show  that  the  value  of  a  function  is  independent  of  the  position 

0       0' 

of  the  perpendicular;  i.e.,  ^  =  ^7,  etc. 

2.  Draw  angles  of  10°,  20°,  30°,  etc.,  to  80°.     In  each  figure  draw  a 
perpendicular  and  measure  the  three  sides  of  the  right  triangle 
formed.     Calculate  the  values  of  the  sine,  cosine,  tangent  and 
cotangent  of  each  angle,  taking  the  results  to  two  decimals. 
Thus,  in  the  figure,  x  =  20°,  0  =  ?,  A  =  ?,  H  =  ?,  sin  20°  = 

^  = =  ?,  cos  20°  =  jr  = =  ?,  tan  20°  =  -v  = =  ?, 

fj[  fj.  A\. 

cot  20°  =£  = =? 


SIMILAR   TRIANGLES   AND    SIMILAR   POYLGONS 
3.  Make  a  table  of  the  results  as  follows : 


171 


Angle 

Sine 

Cosine 

Tangent 

Cotangent 

10° 

.17 

20° 

.34 

30° 

.50    , 

40° 

50° 

60° 

70° 

80° 

Problems  Solved  by  Trigonometry. — A  triangle  whose  sides 
are  many  feet  or  miles  in  length,  or  so  large  that  its  sides 
reach  from  the  earth  to  the  sun  or  from  earth  to  a  star, 
may  be  similar  to  one  of  the  triangles  used  in  calculating 
a  table  of  functions.  The  function  values  of  the  acute 
angles  of  these  larger  triangles  are  therefore  equal  to  the 
values  found  from  the  constructed  triangles. 

EXERCISES 

4.  The  distance  AB  across  a  river  is  required.  A  surveyor  meas- 
ures a  distance  EC  at  a  right  angle  to  the  required  line;  and  at 
point  C  he  measures  Z.C  =  30°.  To  which  of  the  triangles  of 
Exercise  3  is  this  triangle  (full  size)  similar? 


2000  FT. 


A  R 


To  what  function  of  30°  is     TV  equal? 

>  taking  the  proper  value  from 


Form    the   equation,  on?)?) 

the  table.     Solve  for  AB. 

In  solving  the  problems  which  follow,  use  the  more  exact  func- 

tion values  given  in  the  table  at  the  end  of  this  section. 


172  PLANE    GEOMETRY 

5.  The  horizontal  distance  from  a  point  A  on  the  ground,  to  the 
base  of  a  flagpole  is  130  feet.     At  A  the  angle  of  elevation  of 
the  top  of  the  pole  is  50°.     Find  the  height 

of  the  pole  above  the  transit.  Add  4  feet 
for  the  height  of  transit  above  the  ground 
at  the  base  of  the  pole  (Compare  115, 
Exercise  9). 

6.  A  hillside  slopes  at  an  angle  of  20°  with  the 
horizontal.      What  is  the  rise  in  500  feet 
measured  along  the  slope? 

7.  An  aeroplane  is  observed  over  an  aviation  field  that  is  known 
to  be  3.5  miles  distant,  at  an  angle  of  elevation  of  10°.     What 
is  the  altitude  of  the  aeroplane  (Compare  115,  Exercise  4)? 

8.  Find  the  base  of  an  isosceles  triangle  whose  vertex  angle  is  40° 
and  whose  equal  sides  are  20  in. 

Help.  —  Draw  the  bisector  of  the  vertex  angle. 

9.  A  pendulum  10  feet  long  swings  from  end  to  end  of  a  straight 
line  3.4  feet  long.     Through  what  angle  does  the  pendulum 
swing? 

Help.  —  Draw  the  bisector  of  the  total  angle. 

10.  The  radius  of  a  circle  is  10  inches;  radii  drawn  to  the  extremi- 
ties of  a  chord  form  a  central  angle  of  80°.     Find  the  length  of 
the  chord. 

11.  Calculate  the  projection  of  a  sect  5  inches  long  on  a  line  which 
forms  an  angle  of  40°  with  the  sect. 

12.  Calculate  the  angles  of  the  isosceles  triangle  of  97,  Exercise  4. 
Help.  —  Cosine  of  one  of  the  base   Zs  =  %  =  .6667. 

13.  Calculate  the  base  of  the  isosceles  triangle  of  98,  Exercise  3. 


Help.—      ~  =  sine  \  vertex  angle  /.  base  =  6  sine  35°. 
o 

14.  A  vertical  pole  12  feet  high  casts  a  shadow  on  a  horizontal 
plane  13.33  feet  long.     Find  the  angle  of  elevation  of  the  sun. 

15.  The  projection  of  a  sect  20  inches  long  is  17.5  inches.     Find  the 
angle  which  the  sect  makes  with  the  line  of  projection. 

16.  Calculate  the  functions  of  30°,  45°  and  60°  from  the  values 
obtained  in  254. 

Oblique  Triangles  can  also  be  solved  by  dividing  them  into 
two  adjacent  or  overlapping  right  triangles. 

EXERCISES 

17.  The  distance  AB,  across  a  pond  is  required.     A  point  C  is 
selected  from   which  the  lines   CA  =  280.5  feet,   and  CB  = 


SIMILAR   TRIANGLES    AND    SIMILAR    POLYGONS 


173 


620.8  feet  are  measured.     Angle  C 
is  measured  37°. 

Helps.—  (1)  In  AACD,  calculate  p 
and  w;  —  (2)  n  =  a  —  m;  —  (3)  in 
AABD,  calculate  Z.B  and  then  side   C 
c;  or,  calculate  c  =  \/p2  +  n2. 

18.  An  island  C  is  located  on  a  map  by  selecting 
two  known  points  A   and  B  on  the  shore, 
measuring  AB  =  1250  feet,    /.CAB  =  67°; 
^CBA  =  48°.     Find  AC  and  EC. 

Helps.  —  (1)  In  AABD,  calculate  p  and  m; 
—(2)  ZC  =  180°-  (A  +5);—  (3)  in  ABCD, 
calculate  n  and  a;  —  (4)  6  =  m  +  n. 

19.  A  line  is  to  be  extended  past  a 
building.     A  surveyor  measures 
/.DAB  =  153°.  AB  =  325  feet, 
/.ABC  =  105°.      Calculate  BC 

so  that  point  C  shall  be  on  the    D 

line  DA  extended,  and   Z.BCE     -" 

so  that  CE  shall  be  the  prolongation  of  line  DAC. 

20.  The  height  of  a  stack  is  found  as  follows  :  A  transit  is  set  up  at 
point  A    and   the   angle   of 

elevation  of  point  C  is  mea- 

sured,   /.DAC  =  32°;     the 

height  of  the  transit  above 

the  ground  is  measured  on 

the    wall    of    the    building; 

DE  =4.7  feet.      The   tele- 

scope  of  the  transit  is  turned     *- 

over  and  a  point  B  is  located  E 

in  the  plane  of   AACD.     We  will  suppose  that  the    transit 

stands  at  the  same  height  at  B  as  at  A.     Angle  ABC  is  measured 

24°;  AB  is  measured  on  the  ground  =  125.4  feet. 

Helps.—  Calculate  (1)  AF;  (2)   ZACF;  (3)  AC;  (4)  CD;  (5)  CE. 

Solve  the  triangle  of  99,  Exercise  1. 


21. 


Helps.— (1)          =  sine  57°;-(2) 


sine  A; — (3)   -r-^ 


cosine  57°; — (4)  ^-=  =  cosine  A; — (5)  c  =  m  -f-  n; — (6)  C  =  180°  — 
(B  +  A). 

Geometrical  methods  of  calculation  which  are  limited  to 
special  cases  in  geometry,  may  be  generalized  by  trigonome- 
try. 


174  PLANE    GEOMETRY 

EXERCISES 

22.  Find  a  formula  for  calculating  the  projection  m,  of  a  sect  s,  on 
a  given  line  for  any  angle  x,  between  the  sect  and  the  line  of 
projection  (see  254). 

TT      T  Wl 

Helps.  ——=  cosine  x. 
.'.  m   =   s.  cosine  x. 

23.  Calculate  the  projection  of  a  sect  of  24"  which  makes  an  angle 
of  30°  with  the  line  of  projection. 

m  =  24  X  cosine  30°  =  24  X  .8660  =  etc. 

24.  Calculate  the  projection  of  a  sect  of  38"  which  makes  an  angle 
of  51°  with  the  line  of  projection. 

25.  Find  a  general  formula  for  the  third  side  of  a  triangle  when 
two  sides  and  their  included  angle  are  given  (see  252). 

Helps.  —  a2  =  62  +  c2  —  26m;      —  =  cosine  A 
.'.  a2  =  62  +  c2  —  26rcosine  A 

26.  Calculate  the  third  side  of  a  triangle  of  which  two  sides  and 
their  included  angle  are  respectively,  6  =  1.5",  c  =  2.9",  and 
A  =  57°  (see  98,  Exercise  1). 

Helps.—  a*  =  (1.5)2  +  (2.9)2  -  2(1.5)(2.9)  cos  57°. 

27.  Calculate  the  angles  B  and  C  of  the  triangle  of  Exercise  26. 
Helps.  —  (1)  ™  =  cosine  A;  —  (2)n  =  6  -  m;  —  (3)  cosine  C  =  -  ; 
(4)  B  =  180°  -  (A  +  C) 

Some  geometrical  methods  of  calculation  may  be  simplified 
by  trigonometry. 

EXERCISES 

28.  Find  a  formula  for  the  diagonal  of  a  square  in  terms  of  the 
side  s  (260,  Theroem  (14)). 

Helps.-(l)       =  sine  45°;  (2)  d  = 


29.  Calculate  the  diagonal  of  a  square  whose  side  is  50   (261, 
Exercise  41). 

30.  Calculate  the  distance  from  home  plate  to  second  base  in  a 
base  ball  diamond. 

31.  Find  a  formula  for  the  altitude  of  an  equilateral  triangle  in 
terms  of  the  side  of  the  triangle  (260,  Theorem  (16)). 

Help.—  a  =  sine  60°. 
s 

32.  Calculate  the  altitude  of  an  equilateral  triangle  of  which  the 
side  is  120  (261,  Exercise  42). 


SIMILAR    TRIANGLES   AND    SIMILAR    POLYGONS 


175 


TABLE  OF  FUNCTIONS 


Angle 

Sine 

Cosine 

Tangent 

Cotangent 

0° 

0000 

1  0000 

0000 

90 

1° 
2° 
3° 
4° 

.0175 
.0349 
.0523 
.0698 

.9998 
.9994 
.9986 
.9976 

.0175 
.0349 
.0524 
.0699 

57.2900 
28.6363 
19.0811 
14.3007 

89 

88 
87 
86 

5° 
6° 
7° 
8° 
9° 

.0872 
.  1045 
.1219 
.1392 
.1564 

.9962 
.9945 
.9925 
.9903 
.9877 

.0875 
.1051 
.1228 
.1405 

.1584 

11.4301 
9.5144 
8.1443 
7.1154 
6.3138 

85 
84 
83 
82 
81 

10° 
11° 
12° 
13° 
14° 

.1736 
.1908 
.2079 
.2250 
.2419 

.9848 
.9816 
.9781 
.9744 
.9703 

.1763 
.1944 
.2126 
.2309 
.2493 

5.6713 
5.1446 
4.7046 
4.3315 
4.0108 

80 
79 

78 
77 
76 

15° 
16° 
17° 
18° 
19° 

.2588 
.2756 
.2924 
.3090 
.3256 

.9659 
.9613 
.9563 
.9511 
.9455 

.2679 
.2867 
.3057 
.3249 
.3443 

3.7321 
3.4874 
3.2709 
3.0777 
2  .  9042 

75 
74 
73 
72 
71 

20° 
21° 
22° 
23° 
24° 

.3420 
.3584 
.3746 
.3907 
.4067 

.9397 
.9336 
.9272 
.9205 
.9135 

.3640 
.3839 
.4040 
.4245 
.4452 

2.7475 
2.6051 
2.4751 
2.3559 
2.2460 

70 
69 
68 
67 
66 

25° 

26° 
27° 
28° 
29° 

.4226 
.4384 
.4540 
.4695 
.4848 

.9063 
.8988 
.8910 
.8829 
.8746 

.4663 
.4877 
.5095 
.5317 
.5543 

2.1445 
2.0503 
.9626 
.8807 
.8040 

65 
64 
63 
62 
61 

30° 
31° 
32° 
33° 
34° 

.5000 
.5150 
.5299 
.5446 
.5592 

.8660 
.8572 
.8480 
.8387 
.8290 

.5774 
.6009 
.6249 
.6494 
.6745 

.7321 
.6643 
.6003 
.5399 
.4826 

60 
59 
58 
57 
56 

35° 
36° 
37° 
38° 
39° 

.5736 
.5878 
.6018 
.6157 
.6293 

.8192 
.8090 
.7986 
.7880 
.7771 

.7002 
.7265 
.7538 
.7813 
-8098 

.4281 
.3764 
.3270 
.2799 
.2349 

55 
54 
53 
52 
51 

40° 
41° 
42° 
43° 
44° 
45° 

.6428 
.6561 
.6691 
.6820 
.6947 
.7071 

.7660 
.7547 
.7431 
.7314 
.7193 
.7071 

.8391 
.8693 
.9004 
.9325 
.9657 
1.0000 

.1918 
.1504 
.1106 
.0724 
.0355 
.0000 

50 
49 
48 
47 
46 
45 

Cosine 

Sine 

Cotangent 

Tangent 

Angle 

176 


PLANE    GEOMETRY 


Trigonometry  has  been  developed  so  that  it  meets  the 
needs  of  the  most  exact  scientific  work,  such  as  surveying, 
engineering,  navigation,  military  science,  and  astronomy. 
Seven  place  tables  of  the  logarithms  of  the  functions  are 
published,  and  ten  place  tables  are  used  in  some  calculations. 
There  are  also  special  formulas  for  solving  oblique  tri- 
angles, finding  areas,  and  for  many  other  uses.  A  text  on 
trigonometry  should  be  used  if  further  study  is  desired. 

In  using  the  table  on  page  175,  angles  from  0°  to  45°  are 
read  in  the  left  hand  column  and  the  names  of  the  functions 
are  found  at  the  top  of  the  table;  angles  from  45°  to  90°  are 
read  in  the  right  hand  column  and  the  names  of  the  functions 
are  found  at  the  bottom  of  the  table. 

266.  PROBLEMS  FOR  FIELD  WORK. 

1.  The  distance  across  a  stream,  or  inaccessible  street, 
etc.,  by  the  method  of  237. 

2.  The  distance  between  two  points  on  opposite  sides 
of  a  building  or  pond,  by  the  method  of  237. 

3.  Lay  out  a  baseball  diamond  when  the  position  of 
home  plate  and  the  direction  of  second  base  are  given. 

\B 


3 


/\ 

cp/    ^ 
V 


//«- 


Let  H  be  the  home  plate,  and  HC  the  direction  of  center 
field. 

(1)  Find  the  distance  to  second  base,  H2  =  \/902  +  902. 

(2)  Divide  by  2  and  measure  to  point  A. 


SIMILAR   TRIANGLES    AND    SIMILAR    POLYGONS 


177 


B 


(3)  Erect  perpendiculars  A  B,  AD. 

(4)  Measure  Al  =  A3  =  HA  =  ? 

(5)  Check  by  measuring  HI  =  12  =  23  =  3H  =  90  feet. 

4.  Make  a  map  of  some  curved  path  or  road  or  border  of  a 
stream,  by  the  method  of  237. 

5.  Find  the  distance  from  a  position  A  to  a  place  B  as 
follows:  First  observer  proceeds  on  a  line  perpendicular 
to   line   AB,    to    C;    second   observer   proceeds 

along  line  BA  to  D,  until  observer  at  C  signals 
him  that  the  angle  BCD  measured  by  an  optical 
square  (148)  is  a  right  angle.  The  distances  AC 
and  AD  are  paced  or  measured  with  a  tape. 
A  B  may  then  be  calculated  (248). 

This  method  of  estimating  the  distance  to  an 
enemy  station,  etc.,  is  used  in  warfare. 

6.  Find    the    distance    from   a   point   inside 

the  school   grounds  to   an   inaccessible  point  outside  the 
grounds  by  the  method  of  265,  Exercise  4,  or  Exercise  18. 


12 


CHAPTER  XII 
CIRCLES 

PRINCIPLES  DETERMINED  EXPERIMENTALLY 

267.  Circles  Defined. — Refer  to  14  for  general  definitions. 

Congruent,  or  equal,  circles  are  circles  which  have  the 
same  radii. 

Two  points  on  the  circumference  of  a  circle  divide  the 
circumference  into  a  minor  and  a  major  arc. 

A  central  angle  is  formed  by  two  radii.   - 

An  inscribed  angle  is  formed  by  two  chords  meeting  in  the 
circumference. 

.  A  chord  is  said  to  subtend  the  two  arcs  opposite  the  chord, 
and  conversely.  A  central  or  an  inscribed  angle  subtends 
or  intercepts  the  included  arc.  To  subtend  means  to  be  oppo- 
site to. 

A  segment  is  a  part  of  a  circle  bounded  by  a  chord  and  the 
subtended  arc. 

A  sector  is  a  part  of  a  circle  bounded  by  two  radii  and  the 
intercepted  arc. 

An  inscribed  angle  whose  sides  pass  through  the  extremi- 
ties of  a  chord  is  inscribed  in  the  segment  cut  off  by  the  chord. 

Two  circles  may  in  some  positions  have  a  common  chord 
or  common  tangents. 

A  secant  is  a  straight  line  intersecting  a  circle  in  two  points 
and  extending  outside  the  circle  in  one  or  both  directions. 

A  polygon  is  inscribed  in  a  circle  when  all  its  vertices 
lie  in  the  circumference.  The  circle  is  then  circumscribed 
about  the  polygon. 

A  polygon  is  circumscribed  about  a  circle  when  all  its  sides 
are  tangents  of  the  circle.  The  circle  is  inscribed  in  the 
polygon. 

178 


CIRCLES  179 

EXERCISES 

1.  What  part  of  a  circumference  is  an  arc  of  60°;  45°;  25°;  72°  10'? 

2.  Define  complementary,  supplementary  and  explementary  arcs 
(see  17). 

3.  Draw  a  circle  with  a  chord  subtending  an  arc  of  (approximately) 
75°.     What  major  arc  does  the  chord  subtend? 

4.  What  chord  subtends  equal  arcs  of  a  circumference? 

5.  What  must  be  given  in  order  to  determine  the  position  and  size 
of  a  circle? 

6.  Draw  a  circle  with  a  central  angle  subtending  an  arc  of  90°. 
How  many  degrees  are  subtended  by  the  explementary  central 
angle? 

7.  Draw  an  inscribed  angle  subtending  an  arc  of  90°. 

8.  Draw  an  angle  inscribed  in  a  segment  of  (approximately)  150°. 

9.  Define  a  semicircle  as  a  segment. 

10.  Define  a  semicircle  as  a  sector  of  a  certain  number  of  degrees. 

268.  Experiment  I. — Relations  between  central  angles 
and  the  subtended  arcs. 

(a)  Equal  central  angles  in  the  same  or  in  two  equal  circles. 

Draw  a  circle;  draw  several  pairs  of  equal  central  angles. 
Observe  the  relative  lengths  of  the  subtended  arcs  of  each 
pair  of  equal  angles.  State  the  result. 

(6)   Unequal  central  angles  in  the  same  or  in  two  equal  circles. 

Draw  several  unequal  central  angles.  Observe  the  relative 
lengths  of  the  subtended  arcs.  Result. 

(c)  The  measure  of  the  arc  subtended  by  a  central  angle. 
Draw  a  central  right  angle.     Bisect  it;  divide  one  of  the 

parts  (by  trial)  into  three  equal  angles;  divide  one  of  these 
parts  again  into  three  equal  angles;  divide  one  of  these  parts 
into  five  equal  angles.  This  will  give  a  central  angle  of  one 
degree  (of  angle).  Observe  the  part  of  the  circumference 
which  is  subtended  by  the  central  angle  of  one  degree;  the 
number  of  degrees  of  arc  subtended  by  each  central  angle; 
the  number  of  degrees  of  arc  which  will  be  subtended  by 
any  central  angle.  State  result. 

(d)  The  measures  of  the  arcs  subtended  by  a  central  angle 
of  two  concentric  circles. 

Draw  two  concentric  circles;  draw  a  central  angle. 


180  PLANE    GEOMETRY 

Observe  the  relative  lengths  of  the  two  subtended  arcs 
when  measured  in  inches  on  the  circumference;  and  their 
relative  lengths  when  measured  as  fractional  parts  of 
the  two  circumferences,  i.e.,  in  degrees  of  arc.  Result. 

EXERCISES 

1.  What  arc  measured  in  degrees  of  arc  will  a  central  angle  of  75° 
subtend  in  a  circle  of  2"  radius?     In  a  circle  of  5"  radius? 

2.  If  an  arc  of  30°  measures  5"  in  a  given  circle,  what  will  an  arc 
of  90°  measure  in  a  circle  of  double  the  radius  ? 

269.  The  Protractor.— The  principle  of  268(d)  explains 
the  use  of  a  protractor,  which  can  be  used  to  (a)  measure  a 
given  angle;  (b)  construct  an  / 

angle  of  a  given  measure;  (c) 
measure  a  given  arc;  (d)  con- 
struct an  arc  of  a  given  mea- 
sure on  a  given  circumference. 

EXERCISES 

1.  Draw   an   arc   AB  of  about 
one  inch  in  radius.     Measure 
it  in  degrees. 

2.  Construct  an  arc  of  35°  with  a  radius  of  6  inches. 

270.  Experiment  II. — Properties  of  chords. 

(a)  The  relative  lengths  of  arcs  subtended  by  equal  chords, 
in  the  same  or  in  two  equal  circles. 

(b)  The  relative  lengths  of  arcs  subtended  by  unequal  chords. 

(c)  The  relative  distances  (measured  on  perpendiculars)  of 
equal  chords  from  the  center  of  the  circle. 

(d)  The   relative   distances   of  unequal   chords  from   the 
center  of  the  circle. 

(e)  The  relation  of  a  diameter  perpendicular  to  a  chord,  to 
the  chord  and  to  the  two  subtended  arcs. 

(/)  The  property  of  a  chord  equal  in  length  to  the  radius  of 
the  circle. 

In  considering  Part  (6),  observe  whether  the  minor  or 
major  subtended  arcs  vary  in  the  same  way  as  the  chords; 
also  whether  there  is  a  simple  numerical  relation  between 


CIRCLES  181 

the  lengths  of  chords  in  inches  and  the  subtended  arcs  in 
degrees. 

In  considering  Part  (d),  observe  whether  there  is  a 
simple  numerical  relation  between  the  length  of  a  chord 
and  its  distance  from  the  center  of  the  circle. 

In  considering  Part  (/),  draw  a  number  of  chords  equal 
in  length  to  the  radius  of  the  circle,  end  to  end  around  the 
circle. 

State  all  results. 

EXERCISES 

1.  Construct  an  equilateral  triangle  with  its  three  vertices  in  the 
circumference  of  a  circle. 

2.  Construct  a  symmetrical  six-pointed  star  by  drawing  a  circle 
and  using  Part  (/)  above. 

3.  Draw  a  circle  and  lay  off  an  arc  of  120°  by  using  Part  (/)  above 

271.  Experiment  III. — A  method  of 
bisecting  an  arc. 

(a)  By  rule  and  compass. 

(6)  By  measurement. 

(c)  By  trial  and  error. 

Part  (a) :  Use  the  principle  of  270  (e) . 

272.  Experiment  IV. — Properties  of  tangents. 

(a)  The    relative    position    of   a  tangent  and  the  radius 
drawn  to  the  point  of  contact  (or  point  of  tangency). 

(b)  The  relative  lengths  of  two  tangents  drawn  to  a  circle 
from  the  same  external  point. 

(c)  The  number  of  common  tangents  that  can  be  drawn 
to  two  circles  under  different  conditions. 

In  considering  Part  (c),  draw  two  circles  of  different 
radii,  that  are  (1)  externally  out  of  contact,  (2)  externally 
tangent,  (3)  intersecting,  (4)  internally  tangent,  (5)  internally 
out  of  contact.  Observe  the  number  of  common  tangents 
that  can  be  drawn  in  each  case.  State  all  results. 
EXERCISES 

1.  What  position   have  two   tangents   drawn   at  the  ends  of  a 
diameter? 

2.  In  what  position  are  two  circles  if  three  common  tangents  can 
be  drawn  to  them? 


182  PLANE    GEOMETRY 

3.  How  many  common  tangents  have  two  concentric  circles? 

4.  Invent  a  rule  and  compass  construction  for  drawing  a  line 
through  a  given  point  parallel  to  a  given  line,  based  on  a  com- 
mon tangent  to  two  equal  circles. 

5.  Draw  a  circle  and  mark  a  point  on  the  circumference.     Show 
how  the  direction  of  the  tangent  at  this  point  can  be  accurately 
determined  from  a  radius  drawn  to  the  point  of  tangency. 

273.  Experiment  V. — Properties  of  the  line  of  centers 
of  two  circles. 

(a)  The  relation  between  the  line  of  centers  of  two  inter- 
secting circles  and  of  their  common  chord. 

(6)  The  position  of  the  point  of  contact  of  two  tangent 
circles  with  respect  to  the  line  of  centers. 

(c)  The  position  of  the  common  internal  tangent  of  two 
tangent  circles  with  respect  to  the  line  of  centers. 

(d)  The  position  of  the  points  of  intersection  of  a  pair  of 
internal   or   external   common   tangents   with   respect 
to  the  line  of  centers. 

In  considering  Part  (a),  draw  (1)  two  equal  circles,  and 
(2)  two  unequal  circles. 

In  considering  parts  (6)  and  (c),  draw  (1)  two  externally 
tangent  circles,  and  (2)  two  internally  tangent  circles. 

In  considering  part  (d),  draw  two  circles  that  are  (1) 
externally  out  of  contact,  (2)  externally  tangent,  (3) 
intersecting.  State  results. 

EXERCISES 

1.  Draw  a  circle  of  1^"  radius;  mark  a  point  on  the  circumference. 
Draw  a  second  circle  of  2"  radius  tangent  externally  to  the  first 
circle  at  this  point. 

Helps. — Draw  the  radius  of  the  first  circle  to  the  point  of  tangency 
and  extend  it  outside  the  circle.     Use  Part  (6)  above. 

2.  Draw  an  arc  of  3"  radius  and  mark  a  point  on  it.     Draw  a 
second  arc  (or  circle)  of  1"  radius  tangent  internally  to  the  first 
arc  at  the  given  point. 

274.  Experiment  VI. — Properties   of  inscribed   angles. 

(a)  The  size  of  an  angle  inscribed  in  a  semicircle. 

(b)  The  size  of  an  angle  inscribed  in  a  minor  segment. 


CIRCLES  183 

(c)  The  size  of  an  angle  inscribed  in  a  major  segment. 

(d)  The  relative  size  of  angles  inscribed  in  the  same  segment. 

(e)  The  measure  of  an  inscribed  angle  in  terms  of  the 
subtended  arc. 

State  results. 

275.  EXPERIMENT   VII. — A  method  of  drawing  a  line 
perpendicular  to  a  given  line. 

(a)  When  the  point  is  in  the  given  line. 

(b)  When  the  point  is  without  the  given  line. 
Adapt  Part  (a)  of  274  to  both  constructions. 

276.  EXPERIMENT   VIII. — The  number  of  circles  that 
can  be  drawn  through  given  points. 

(a)  One  given  point. 

(b)  Two  given  points. 

(c)  Three  given  points. 

(d)  Four  or  more  given  points. 

(e)  The  conditions  that  four  or  more  given  points  shall 
lie  on  the  same  circle. 

In  stating  the  results  of  Parts  (a),  (6),  (c),  state  where  the 
centers  of  the  circles  which  contain  the  given  points  are 
found,  as  well  as  the  number  of  such  circles ;  that  is,  describe 
the  locus  of  their  centers. 

Part  (d) :  If  the  points  are  placed  at  random,  will  any 
circle  contain  them  all?  RESULT. — In  general,  etc. 

Part  (e):  A  certain  set  of  lines  must  be  concurrent  in 
order  that  the  points  shall  be  on  a  circle  (i.e.,  concyclic). 

EXERCISES 

1.  What  is  the  locus  of  the  centers  of  circles  which  contain  the  ends 
of  a  sect? 

2.  Can  a  circle  be  drawn  containing  the  vertices  of  a  square?     Of 
an  oblong?     Of  a  rhombus?     Of  an  isosceles  trapezoid? 

3.  Are  the  vertices  of  an  oblique  scalene  triangle  concyclic? 

4.  How  many  points  may  be  concyclic?     How  many  points  must 
be  concyclic? 

5.  If  three  points  chance  to  lie  in  the  same  straight  line,  where  is 
the  center  of  the  circle  which  may  be  said  to  contain  them? 


184  PLANE    GEOMETRY 

277.  EXPERIMENT  IX. — THE  NUMBER  OF  CIRCLES  THAT 
CAN  BE  DRAWN  TANGENT  TO  GIVEN  STRAIGHT  LINES. 

(a)  One  given  line. 

(b)  Two  given  lines. 

(c)  Three  given  lines. 

(d)  Four  or  more  given  lines. 

(e)  The  condition  that  four  or  more  given  straight  lines 
shall  be  tangent  to  the  same  circle. 

Part  (6):  Consider  two  cases,  (1)  the  given  lines  inter- 
secting, (2)  the  lines  parallel. 

Part  (c):  Consider  the  lines,  (1)  two  lines  parallel  and 
the  third  line  intersecting  them,  (2)  the  three  lines  parallel, 


(3)  the  lines  inclosing  a  triangle,  (4)  the  lines  concurrent. 
In  (3),  note  that  four  circles  can  be  drawn  tangent  to  the 
three  given  lines. 

Part  (e):  Find  the  set  of  lines  that  must  be  concurrent 
as  the  necessary  condition. 

EXERCISES 

1.  What  is  the  locus  of  the  centers  of  the  circles  in  Part  (b),  (1) 
and  (2)? 

2.  What  are  the  tangent  circles  called  in  respect  to  the  triangle 
enclosed  by  the  three  lines  in  Part  (c),  (3)? 

3.  Can  a  circle  be  drawn  tangent  to  the  sides  of  a  square?     Of  an 
oblong?     Of  a  rhombus?     Of  an  isosceles  trapezoid? 

278.  EXPERIMENT     X. — ANY     OTHER    PROPERTIES    OF 
CIRCLES  WHICH  CAN  BE  DISCOVERED  EXPERIMENTALLY. 

279.  Construction  of  Circles  from  Given  Data. — Find 
all  possible  solutions  of  each  problem.     Discuss  special  and 
impossible  cases,  which  should  be  illustrated  by  sketches. 


CIRCLES 


185 


Use  rule  and  compass  or  measurement  methods,  and  avoid 
trial  and  error  solutions. 

Construct  a  circle  (or  circles)  of  which  there  are  given: 
1.  The  length  and  position  of  a  chord,  and  the  radius. 


B 


Helps. — AB  is  the  given  chord;  r  is  the  given  radius. 

In  general  two  circles  can  be  drawn. 

Special  cases:  (1)  When  r<%  AB,  no  circle  is  possible.     (2) 

When  r  =  ^  AB,  one  circle,  as  C,  can  be  drawn. 

2.  The  length  and  position  of  a 
chord  and  the  position  of  a  tangent-^" 

through  an  end  of  the  chord. 

3.  (The  position  of)  a  tangent, 
the  point  of  tangency,  and  a 
point  on  the  circumference. 

4.  Two    intersecting    tangents 
and  the  radius. 

5.  A  chord  and  a  straight  line  containing  the  center  of  the  circle. 

6.  A  tangent   circle  and  the  center  of  the 
required  circle. 

7.  Two    non-intersecting    circles,    both    of 

which  are  tangent  to  the  required  circle,     \         ^  /         x 

and  the  radius  of  the  required*  circle.  \  /          B 

8.  A  tangent  circle  and  a  point  in  the  cir- 
cumference of   the   required  circle  and  the  radius  of  the  re- 
quired circle. 

9.  The  length  and  position  of  a  chord  and  a  circumference  con- 
taining the  center  of  the  required  circle. 

10.  A  tangent  circle  and  the  point  of  tangency,  and  a  point  in  the 
circumference  of  the  required  circle. 


186 


PLANE    GEOMETRY 


11.  State  other  conditions  for  the  construction  of  a  required  circle 
(or  circles).     Make  the  conditions  definite  but  not  conflicting. 


280.  Constructions  Relating  to 

rule  and  compass  methods: 


Circles. — Construct  by 


1.  A  tangent  to  a  given  circle  parallel  to  a 
given  straight  line. 

2.  A  tangent  to  a  given  circle  perpendicular 
to  a  given  straight  line. 

3.  A  chord  in  a  given  circle  of  given  length 
and  parallel  to  a  given  straight  line. 

4.  A  chord  of  a  given  circle  passing  through  a  given  point  within 
the  circle  and  bisected  at  that  point. 

5.  A    right    triangle     when     the     ^  H 

lengths  of  the  hypotenuse  and      p   {         

of  one  of  the  perpendicular 
sides  are  given.  Construct  (1) 
by  274 (a);  (2)  by  any  other 
method. 

6.  A    right    triangle    when    the 
hypotenuse  and  the  altitude  to 
the  hypotenuse  are  given. 

7.  A  chord  of  given  length  in  a 

given  circle,  which  shall^pass^through  a  given  point  within 
the  circle. 

8.  Four  circles  within  a  given  square,  each  tangent  to  the  sides  of 
the  square  and  to  the  two  adjacent  circles. 

9.  A  chord  of  given  length  in  a  given  circle,  which  shall  pass  (if 
produced)  through  a  given  point  outside  the  circle. 

281.  REVIEW  EXERCISES 

1.  Draw  two  tangent  arcs  of  radii  2"  and  3.5",  45°  and  60°  respec- 
tively, in  two  positions  of  contact. 

2.  Construct   the   locus   of 
the     centers    of    circles 
tangent  to  a  given  circle 
at  a  given  point  on  the 
circle. 

3.  Construct  an  arc  of  60°, 

6"  radius,  without  using  a  protractor. 

4.  Construct  a  circle  (or  circles)  tangent  to  two  given  intersecting 
straight  lines  and  having  its  center  in  a  third  given  straight  line. 

5.  Construct  a  circle  tangent  to  a  given  circle  and  to  a  given 
straight  line  and  having  a  given  radius. 


CIRCLES  187 

6.  Two  chords  intersect  on  the  circumference  of  a  circle  and  have 
their  other  ends  at  the  ends  of  a  diameter  of  the  circle.     The 
chords  are  5"  and  6"  long.     Calculate  the  diameter. 

7.  A  chord  of  a  circle  is  10"  and  its  distance  from  the  center  is  6". 
Calculate  the  radius  of  the  circle. 

8.  The  perpendicular  from  the  center  of  a  circle  to  a  chord  sub- 
tending an  arc  of  120°  is  3".     Calculate  the  radius  and  the 
length  of  the  chord. 

9.  A  chord  subtending  an  arc  of  60°  is  5"  from  the  center.     Calcu- 
late the  radius  and  the  chord. 

10.  The  radius  of  a  circle  is  10".     Calculate  the  length  of  a  chord 
subtending  an  inscribed  angle  of  45°. 

11.  The  center  sect  of  two  circles  is  10";  the  radii  are  4"  and  2". 
Calculate  the  length  of  an  internal  common  tangent. 

12.  Draw  two  externally  tangent  circles.     Prove  experimentally 
that  the  internal  common  tangent  bisects  the  two  external 
common  tangents. 

13.  Draw  a  number  of  chords  of  a  circle  concurrent  in  a  point  of 
the  circumference.     What  is  the  locus  of  their  middle  points? 

14.  Draw  a  number  of  secants  of  a  circle  concurrent  in  a  point  out- 
side the  circle.     What  is  the  locus  of  the  bisection  points  of 
both  the  major  and  minor  segments  of  the  secants? 

15.  Make  an  experiment  similar  to  that  of  Exercise  14  for  a  pencil 
of  chords  intersecting  within  the  circle. 

16.  Draw  a  circle  or  an  arc  of  more  than  180°  without  marking 
the  center.     Find  the  center  by  using  the  right  angle  of  the 
draughtsman's  triangle  (274  (a)). 

17.  Construct  a  circle  tangent  6 
to  the  two  circles  A  and 

B  and  containing  the 
point  C.  Draw  (a)  the 
locus  of  points  equidis- 
tant from  circle  A  and 
point  C;  (6)  a  similar 
locus  for  circle  B  and 
point  C;  (c)  the  required  circle  with  center  at  O. 

18.  Exercise  17  is  the  geometric  solution  of  the  problem  of  locating 
the  German  " super-gun"  which  shelled  Paris  in  1917-18.     See 
Mathematics  Teacher  for  Dec.,  1918. 

Three  stations,  A,  B,  C,  were  located  and  plotted.  A  listening 
apparatus  was  installed  at  C,  and  observers  were  stationed  at 
A  and  B.  Each  observer  timed  with  accurate  stop  watches 
the  interval  between  hearing  the  discharge  of  the  gun  at  C  and 
hearing  it  at  his  station,  A  or  B.  Let  this  interval  be  1.7 


188  PLANE    GEOMETRY 

seconds  for  the  A  observer,  and  2.5  seconds  for  the  B  observer. 
Sound  travels  about  1100  feet  per  second.  Draw  the  circle  A 
with  a  radius  1.7  X  1100  feet,  and  circle  B  with  radius  2.5  X 
1100  feet.  Circle  O  is  found  as  in  Exercise  17,  or  by  trial;  at 
the  center,  O,  of  which  the  gun  is  located. 

APPLICATIONS 

282.  Railroad  and  Road  Curves. — A  preliminary  survey 
of  a  proposed  railroad  consists  in  staking  out  a  series  of 
straight  lines,  locating  the  points  of  intersection,  measuring 
the  lengths,  AB,  BC,  etc.,  (Fig.  1)  and  the  inter  section  angles, 
x,  y,  etc.;  also  in  estimating  the  radii  of  the  connecting 
circular  arcs  which  will  best  conform  to  the  surface  of  the 
ground. 

The  lines  AB,  BC,  etc.,  are  known  as  tangents. 


FIG.  1. 
EXERCISES 

1.  Show  that  the  central  angle  at  O  equals  the  intersection  angle  x 
(Fig.  2). 


FIG.  2. 


CIRCLES 


189 


2.  How  do  the  tangent  lengths  BH  and  BK  compare  (Fig.  2)? 

3.  Plot  the  tangents  on  a  scale  of  1000  feet  to  the  inch,  from  the 
following  notes  (Fig.  1): 


Tangents 

Lengths, 
feet 

Intersection 
IS. 

Values 

Direction 
of  turn 

AB 

2000 

B 

40° 

To  right 

BC  
CD  
DE  

2500 
1500 
3000 

C 
D 
E 

90° 
50° 

35° 

To  left 
To  right 
To  right 

To  Lay  Out  a  Circular  Arc  by  Chords. — The  necessary 
calculation  are  made  by  trigonometry.  Thus  if,  in  Fig.  2, 
x  =  40°,  £HOK  =  40°,  and  the  arc  HK  may  be  divided 
into  10  equal  arcs  of  4°  each.  The  radius  HO  =  1433  feet, 
the  tangents,  BH  =  BK  =  522  feet;  and  the  chords  HL  = 
LM  =  MN  =  100  feet.  Set  up  a  transit  at  H,  sight  on 
B,  turn  2°  and  measure  100  feet  to  L;  turn  another  2°  and 
measure  LM  =  100  feet;  etc. 


EXERCISES 

4.  Plot  the  curve  described  above  on  the  drawing  of  Exercise  3. 

5.  Why  are  the  angles  BHL  =  LHM  =  MHN  =  etc.  =  2°  (Fig. 

2)? 


The  Two -transit  Method  of  Laying  Out  a  Circular  Arc.— 

The  arc  HK  may  also  be  laid  out  with  two  transits,  without 
making  any  measurements  with  a  tape. 

After  calculating  BH  and  BK  (Fig.  2)  and  locating  points 
H  and  K,  set  a  transit  at  each  of  these  points.  In  order  to 
locate  points  L,  M,  N,  etc.;  turn  ZBHL  =  2°  with  the  H- 
transit,  and  an  angle  BKL  =  18°  with  the  X-transit,  and  set 
a  stake  on  both  lines  of  sight;  then  turn  /.BUM  =  4°  and 
£BKM  =  16°,  and  set  stake  M  as  before;  etc. 


190 


PLANE   GEOMETRY 


EXERCISE 

6.  Arrange  the  angles  of  each  transit  in  a  table: 


Point  on  arc 

Z  turned  by  //-transit 

Z  turned  by  /^-transit 

L 

2° 

18° 

M 

N 

Etc. 

Compound  Curves. — Such  curves  are  composed  of  two 
or  more  arcs  successively  tangent  to  each  other. 

A  reverse  curve  is  a  compound  curve  composed  of  two 
arcs  of  equal  radii,  joining  two  parallel  tangents. 


FIG.  1. 


FIG.  2. 


EXERCISES 


7.  Design,  by  trial,  a  compound  curve  in   which  arc  AB  (Fig.l) 
has  a  radius  of  300  feet  and  arc  BC  a  radius  of  500  feet,  and  in 
which  the  intersection  angle  x  =  90°. 

8.  Design  a  reverse  curve  joining  two  parallel  tangents  the  distance 
between  which  is  30  feet,  the  radius  of  both  arcs  being  30  feet. 

£)2     I      |^2 

9.  Show  that  in  Fig.  2,  R   =   —  ~  —  . 


Help.—OiE  =  2R  -W't  (hO~22  =  Ch 

10.  Calculate  R  when  D  =  W;  also  when  D  =  W  =  30  ft.;  also 
when  D  =  30  ft.  and  W  =  20  ft. 


CIRCLES 


191 


Y-curves. — Such  curves  connect  two  railroad  tracks  as 
shown,  and  are  sometimes  used  to  reverse  the  direction  of 
an  entire  train. 


EXERCISES 

11.  Show  that  EC  =  EF  =  ED. 

12.  Show  that  EC  =  <\/frr,  where  R  and  r  are  the  radii  of  the  curves. 
Helps. — AAEB  is  a  right  A.     Apply  248. 

13.  Calculate  EC  when  R  =  r  =  200  ft.     Also  when  R  =  200  ft. 
and  r  =  150  ft. 

283.  THE  CIRCLE  IN  ARCHITECTURE. — Arches,  columns, 
domes,  doorways,  windows,  and  other  architectural  features 
exhibit  many  beautiful  and  interesting  uses  of  the  circle. 

The  Gothic  Arch. — A B  is  the  span;  CD  is  the  altitude. 

EXERCISES 

1.  Design  a  Gothic  arch  with  as  pan  of  40  ft.,  and  an  altitude  of 
30ft. 

2.  Design  a  Gothic  arch  with  a  span  of  30  ft.,  and  an  altitude  of 
40ft. 


The  Breccia  Arch. — A  ABC  is  equilateral. 
3.  Design  a  Breccia  arch  with  a  span  of  40  ft.  and  an  altitude  DE 
of  30  ft. 


192 


PLANE    GEOMETRY 


The  Persian  (ogee)  Arch. — Trisect  AC  at  E  and  F;  draw 
FG±AC;  prolong  GE  to  H;  G  and  H  are  the  arc  centers. 
The  arch  may  also  be  drawn  with  E  at  the  middle  of  AC. 

4.  Design  Persian  arches  with  30  ft.  span  and  40  ft.  altitude;  (a) 
with  E  at  the  trisection  point;  (6)  with  E  at  the  bisection  point. 


The  Basket-handle  Arch. 

5.  Design  a  basket-handle  arch  whose  span  =  60  ft.,  altitude  = 
25ft. 

The  Tudor  Arch. 

6.  Design  a  Tudor  arch   whose  span  =  60  ft.,  altitude  =  30  ft. 


Circular  Arches. 

7.  Design  circular  arches:  (a)  span    =  120  ft.,  altitude 
(6)  span  =  120  ft.,  altitude  =  40  ft. 


60  ft.; 


CIRCLES 


193 


284.  A  PROBLEM  FOR  FIELD  WORK. — To  lay  out  a  circu- 
lar arc  tangent  to  two  straight  lines,  by  the  two-transit 


method  (282).  Select  two  straight  curbs  or  center  lines  of 
paths,  and  mark  such  points  as  A,  B,  C,  D,  by  driving  stakes 
or  with  a  chisel.  Set  transits  over  points  A  and  D  and  sight 
on  points  B  and  C  respectively,  and  set  stake  E  on  the  two 
lines  of  sight.  Set  the  transit  at  E  and  measure  angle 
AED.  Measure  any  equal  distances  EF  and  EG,  say  200 
feet.  The  central  angle  0  of  the  arc  connecting  points 
F  and  G  =  180°  -  LAED  =  Z.FOG.  Divide  Z.FOG  by 
10  =  /.FOH.  Set  the  transits  over  points  F  and  G.  Sight 
the  /^-transit  on  point  G  and  turn  an  angle  GFH  =  9 
times  |  /.FOH.  Sight  the  (7-transit  on  point  F  and  turn 
an  angle  FGH  =  \£FOH.  Set  stake  H  on  the  line  of 
sight  of  both  transits.  Proceed  in  this  way  until  the  9 
stakes  H,  7,  J,  K,  etc.,  are  located. 

This  problem  is  greatly  simplified  by  selecting  two  perpen- 
dicular lines,  or  by  laying  out  two  perpendicular  lines  from 
any  selected  point  E. 


13 


CHAPTER  XIII 
CIRCLES 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

285.  The  first  step  in  classification  is  the  selection  of 
the  most  elementary  experimental  properties  of  circles  as  a 
basis  of  analysis. 

CIRCLE  POSTULATES 

1.  Any  point  within  a  circle  is  at  a  less  distance  from  the 
center  than  the  radius;  and  conversely. 

2.  Any  point  without  a  circle  is  at  a  greater  distance  from 
the  center  than  the  radius;  and  conversely. 

3.  If  two  or  more  points  are  equally  distant  from  another 
point,  a  circle  may  be  drawn  with  its  center  at  that 
other  point  which  will  contain  all  of  the  equidistant 
points. 

4.  A  circle  is  determined  by  (a)  the  center  and  radius; 

(6)  by  a  diameter. 

5.  A  diameter  bisects  a  circle. 

6.  Equal  chords  of  the  same  circle  or  of  equal  circles 
subtend  equal  arcs;  and  conversely. 

7.  Equal  chords  of  the  same  circle  or  of  equal  circles  are 
equally  distant  from  the  center;  and  conversely. 

8.  Unequal  chords  of  the  same  circle  or  of  equal  circles 
subtend  unequal  arcs;  the  greater  chord  subtending 
the  greater  minor  arc. 

9.  Unequal  chords  of  the  same  circle  or  of  equal  circles 
are  unequally  distant  from  the   center,  the  greater 
chord  being  at  the  less  distance  from  the  center. 

10.  A  diameter  is  the  greatest  chord. 

194 


CIRCLES  195 

11.  (a)  A  tangent  is  perpendicular  to  the  radius  drawn 
to  the  point  of  tangency.     (6)  A  line  perpendicular 
to  a  radius  at  its  extremity  is  tangent  to  the  circle. 

12.  (a)  Every  point  of  a  tangent  lies  outside  the  circle 
except  the  point  of  contact.     (6)  If  every  point  of 
a  straight  line  lies  outside  a  circle  except  one  point, 
the  line  is  a  tangent. 

13.  Equal  central  angles  of  the  same  circle  or  of  equal 
circles  subtend  equal  arcs;  and  conversely. 

14.  A  central  angle  has  the  same  measure  in  degrees 
of  angle  as  the  subtended  arc  has  in  degrees  of  arc. 
Or,  a  central  angle  equals  its  subtended  arc. 

15.  The  line  of  centers  of  two  tangent  circles  contains  the 
point  of  tangency. 

286.  The  second  step  in  classification  is  the  analysis  of 
other  experimental  properties  of  circles  in  terms  of  more 
elementary  properties. 

287.  Theorem  I. — In  the  same  circle  or  in  equal  circles, 
equal  chords  subtend  equal  central  angles; 

and  conversely. 

Parti. 

Hypothesis. —  Circle  0  with  equal 
chords  AB  and  CD,  subtending  central 
angles  x  and  y. 

Conclusion. — x  =  y. 

Analysis. 

1.  Arc  AB  =  arc  CD  What  postulate? 

2.  .'.  /.x  =  /.y  What  postulate  ? 
Part  II. 

Hypothesis. — Circle  0  with  equal  central  angles  x  and  y, 
subtending  chords  AB  and  CD. 

Conclusion. — Chord  AB  =  chord  CD. 
Analysis. 

1.  Arc  AB  =  arc  CD  What  postulate? 

2.  .*.  chord  AB  =  chord  CD          What  postulate? 
Help. — The  postulates  used  are  6  and  converse,  and  13 

and  converse. 


196 


PLANE    GEOMETRY 


288.  Theorem  II. — The  perpendicular  bisector  of  a  chord 
contains  the  center  of  the  circle. 

Help. — Use  a  locus  theorem  or  a  theorem  of  isosceles 
triangles.  (See  also  217,  Theorem  (5) .) 

Corollaries. —  (1).  The  perpendicular  bisectors  of  two 
non-parallel  chords  of  a  circle  determine  the  center  of 
the  circle. 

2.  Three  points  on  the  circumference  of  a  circle  determine 
the  center  of  the  circle. 

289.  Theorem  III. — A  diameter  perpendicular  to  a  chord 
bisects  it  and  both  major  and  minor  subtended  arcs. 

Corollary. — A  radius  perpendicular  to  a  chord  bisects 
it  and  the  subtended  arc. 

290.  Theorem  IV.— Two  tangent 
circles  have  a  common  tangent  at  the 
point  of  tangency. 

Helps. — (1)    Draw    AB,    which 
contains  point  C;   Postulate?     (2) 
Draw  CD  tangent  to  circle  A;   (3) 
CD.LAC;  why?     (4)  /.   CD  is  tangent  to  circle  B;  why? 
Analyze  also  for  two  internally  tangent  circles. 

291.  Theorem  V. — A  perpendicular  to  a  A          B 
tangent  at  the  point  of  tangency  contains  the 

center  of  the  circle. 

Helps.— (1)  Draw  OA;  (2)  OA  coincides 
with  AC;  etc. 

292.  Theorem  VI. — Two  tangents  drawn 
to  a  circle  from  an  exterior  point  are  equal. 

Helps. — (a)  Draw  a  line  from  the  point  to  the  center  of 
the  circle,  and  radii  to  the  points  of  tangency.  Prove 
that  the  triangles  are  congruent.  Or,  (6)  connect  the 
points  of  tangency  and  draw  radii  as  before.  Use  principles 
of  isosceles  triangles.  Or,  (c)  draw  a  circle  concentric  to 
the  given  circle  and  containing  the  given  exterior  point. 
Prolong  the  tangents  so  that  they  shall  be  chords  of  this 
larger  circle.  Use  circle  Postulate  11  (a)  and  7  converse, 
and  Theorem  III  or  Corollary. 


CIRCLES 


197 


ADDITIONAL  THEOREMS 

293.  (1)  The  line  of  centers  of  two  intersecting  circles  is 
a  perpendicular  bisector  of  the  common  chord  (See  also 
217,  Theorem  (6).) 

(2)  The  line  which  joins  the  center  of  a  circle  with  the  center 
of  a  chord  is  perpendicular  to  the  chord  and  bisects  the  arc 
subtended  by  the  chord. 

(3)  A  radius  which  bisects  an  arc  bisects  the  chord  sub- 
tended by  the  arc  and  is  perpendicular  to 

the  chord. 

(4)  //  a  straight  line  intersects  two  con- 
centric circles  the  two  sects  intercepted  between 
the  circles  are  equal. 

(5)  Two   chords  which  are  both  perpen- 
dicular to  the  same  chord  are  equal. 

Help.— Show  that  OA  =  OB. 

(6)  Two    equal    chords    which    intersect 
within  a  circle  divide  each  other  so  that  the 
segments  of  one  chord  equal  the  segments  of 
the  other. 

(7)  A  line  drawn  from  the  center  of  the 
circle    to    the    point    of  intersection  of  two 
equal  intersecting  chords,  bisects  the  angle 
formed  by  the  chords. 

Analyze     also     when    the     chords     intersect 
the  circle. 

(8)  //  two  secants  which  intersect  outside  a  circle  make 
equal  angles  with  a  line  drawn  from  the  center  of  the  circle  to 
the  point  of  intersection  of  the  secants,  the  whole  secants  are 
equal  and  the  segments  into  which  they  are  divided  by  the  circle 
are  respectively  equal. 

(9)  A  side  of  an  equilateral  inscribed  hexagon  is  equal  to 
the  radius  of  the  circle. 

Helps. — Draw  radii;  the  central  angles  are  equal;  each 
central  angle  is  60°;  each  triangle  formed  is  equiangular; 
.'.  each  triangle  is  equilateral. 

(10)  Two  equal  chords  which  intersect  within  a  circle  inter- 
cept one  pair  of  equal  opposite  arcs. 


without 


198 


PLANE    GEOMETRY 


(11)  Two  chords  which  make  equal  angles  with  a  diameter 
drawn  through  their  point  of  intersection  are  equal. 

(12)  A  chord  which  is  drawn  through  a 
given  point  within  a  circle  perpendicular 
to  the  diameter  through  the  given  point, 
is   shorter  than  any  other  chord  drawn 
through  the  given  point. 

Helps.  —  Draw    OB;    show  that 
OA>OB;etc. 

(13)  A  chord  of  the  larger  of  two  con- 
centric circles   which   is    tangent    to  the 
smaller  circle  is  bisected  at  the  point  of 
tangency. 

(14)  The    line   joining    the    point    of 
intersection  of  two  tangents  with  the  center 

of  the  circle:  (a)  bisects  the  angle  formed  by  the  tangents; 
(b)  bisects  the  central  angle  formed  by  the  radii  to  the 
points  of  tangency;  (c)  bisects  the  arcs  subtended  by  the 
tangents;  (d)  is  a  perpendicular  bisector  of  the  chord  of  contact 
of  the  tangents. 

(15)  The  side  of  a  circumscribed  square  equals  the  diameter 
of  the  circle. 

(16)  The  tangents  drawn  at  the  extremities  of  a  diameter  are 
parallel. 

(17)  The  tangents  drawn  at  the  extremities 
of  a  chord  make  equal  angles  with  the  chord. 

(18)  Chords  of  the  larger  of  two  concentric 
circles  which  are  tangent  to  the  smaller  of  the 
two  circles,  are  equal. 

(19)  The  sum  of  two  opposite  sides  of  a 
circumscribed  quadrilateral  equals  the  sum  of  the  other  two 
opposite  sides. 

Help. — Apply  292. 

(20)  The  central  angle  formed  by  radii  which  are  drawn  to 
the  points  of  tangency  of  two  tangents  which  intersect  outside 
the  circle,  is  supplementary  to  the  angle  formed  by  the  tangents. 

(21)  The  line  of  centers  of  two  circles  externally  out  of  con- 


CIRCLES  199 

tact  contains  the  points  of  intersection  (a)  of  the  internal 
common  tangents,  and  (6)  of  the  external  common  tangents. 
Helps. — Draw  lines  from  the  point  of  intersection  of  the 
tangents  to  the  centers  of  the  circles.  Show  (a)  that  these 
lines  lie  in  the  same  straight  line;  (b)  that  they  coincide. 

294.  The  Third  Step  in  Classification. — The  more  impor- 
tant experimental  results  of  Chapter  XII  have  been  classi- 
fied, but  there  are  many  measurement  properties  of  angles 
and  sects  associated  with  circles,  which  are  not  evident 
from  the  inspection  of  a  constructed  figure  and  can  only 
become  known  by  extending  the  methods  of  analysis  into 
the  realms  of  discovery. 

295.  Angle  Properties  of  Lines  Drawn  to  a  Circle. — Any 
two  intersecting  lines  which  intersect  or  meet  or  touch  a 
circle,  form  an  angle  whose  measure  is  related  to  the  measure 
of  the  subtended  or  intercepted  arc  or  arcs.     The  simplest 
of  these  angle  measures  are  stated  in  Postulates  13  and  14. 
Other  similar  properties  will  now  be  deduced. 

296.  Theorem    VII. — An    inscribed    angle    has    one-half 
the  measure  of  the  intercepted  arc;  or,  it  is  measured  by  one- 
half  the  intercepted  arc;  or,  it  equals  one-half,  etc. 

This  principle  is  best  considered  in  three  cases. 
Case  1. — When  one  side  of  the  inscribed  angle  is  a  diameter. 

6 


Hypothesis. — Inscribed  /.x,  AB  being  a  diameter. 
Conclusion. — Zz  =  ^  arc  AC. 
Analysis. — (1)  Draw  OC. 

(2)  y  =  arc  AC 

(3)  x    =  %y. 

(4)  .'.  x  =  etc. 


200 


PLANE    GEOMETRY 


Case  2. — When  the  center  of  the  circle  is  within  the  inscribed 
angle. 

3 


Analysis. — (1)  Draw  diameter  BD. 

(2)  y  =  \  arc  AD. 

(3)  z   =  \  arc  DC. 

(4)  .*.  x  =  etc. 

Case  3. — State  and  analyze  this  case. 
Corollaries. — (1)  Angles  inscribed  in  the  same   segment 
are  equal. 

2.  Inscribed  angles  subtending  equal  arcs  are  equal. 

3.  An  angle  inscribed  in  a  semicircle  is  a  right   angle. 
(See  also  143,  Theorem  (7).) 

4.  A  principle  relating  to  an  angle  inscribed  in  a  major 
or  a  minor  segment. 

EXERCISES 

1.  What  inscribed  angle  intercepts  an  arc  of  95°? 

2.  What  arc  does  an  inscribed  angle  of  95°  intercept? 

297.  Theorem  VIII. — The  angle  formed  by  a  tangent  and 
a  chord  drawn  from  the  point  of  tangency  has  a  measure  in 
terms  of  subtended  arcs. 

Helps. — There  are  two  of  these  angles.  Draw  a  diameter 
through  the  point  of  tangency;  one  of  the  angles  to  be  con- 
sidered is  the  sum,  and  the  other  is  the  difference,  of  a  right 
angle  and  an  inscribed  angle.  The  measures  of  these  angles 
in  terms  of  arcs  can  be  expressed  from  known  principles. 
State  the  theorem  discovered. 


CIRCLES  201 

EXERCISES 

1.  A  chord  subtends  an  arc  of  80°,  and  a  tangent  is  drawn  at  one 
extremity  of  the  chord.     Calculate  the  value  of  the  angle  formed. 

2.  A  chord  meets  a  tangent  at  the  point  of  tangency  at  an  angle  of 
75°.     What  is  the  value  of  the  arc  subtended  by  the  chord  (or 
intercepted  by  the  angle)  ? 

3.  A  chord  subtends  an  arc  of  100°,  and  tangents  are  drawn  at  each 
end  of  the  chord.     What  angle  is  formed  by  the  tangents  at 
their  point  of  intersection? 

298.  Theorem  IX. — The  angle  formed  by  two  chords  which 
intersect  within  a  circle  has  a  measure  in  terms  of  subtended 
arcs. 

Helps. — Draw  a  chord  connecting  extremities  of  the  given 
chords,  forming  a  triangle;  the  angle  to  be  considered  is 
related  to  angles  of  this  triangle,  which  are  measured  by 
certain  arcs  as  previously  analyzed.  Draw  the  elementary 
figures.  State  the  theorem. 

EXERCISES 

1.  Two  chords  which  intersect  within  a  circle  intercept  opposite 
arcs  of  70°  and  100°.     What  angle  do  the  chords  form? 

2.  Two  chords  intersect  within  a  circle  forming  an  angle  of  30°; 
one  of  the  two  intercepted  opposite  arcs  is  20°.     What  is  the 
value  of  the  other  of  the  opposite  arcs? 

299.  Theorem  X. — The  angle  formed  by  two  secants  which 
intersect  outside  the  circle  has  a  measure  in  terms  of  the  two 
intercepted  arcs. 


Helps. — Draw  a  chord  as  shown.  The  angle  to  be  con- 
sidered has  a  relation  in  terms  of  two  inscribed  jangles. 
Draw  the  elementary  figures.  Analyze  also  by  drawing  a 
different  chord  from  that  shown  in  the  figure. 


202 


PLANE    GEOMETRY 


300.  Theorem  XI. — The  angle  formed  by  a  secant  and 
a  tangent  which  intersect  outside  the  circle  has  a  measure  in 
terms  of  the  two  intercepted  arcs. 

301.  Theorem  XII. — The  angle  formed  by  two  tangents 
which  intersect  outside  a  circle  has  a  measure  in  terms  of  the 
two  intercepted  arcs. 

Corollary. —  The  angle  formed  by  two  tangents  which  inter- 
sect outside  a  circle  has  a  measure  in  terms  of  the  smaller 
intercepted  arc  alone. 

Helps.— (1)  A  =  J(*  -  2/);  (2)  x  +  y  =  360°;  (3) 
eliminate  x. 

EXERCISES 

1.  Two  secants  which  intersect  without  a  circle  intercept  arcs  of 
80°  and  40°.     What  angle  do  the  secants  form? 

2.  A  secant  and  a  tangent  which  intersect  without  a  circle  intercept 
arcs  of  100°  and  40°.     What  angle  do  the  secant  and  tangent 
form? 

3.  Two  tangents  which  are  drawn  from  the  same  exterior  point 
intercept  arcs  of  200°  and  160°.     What  angle  do  the  tangents 
form? 

4.  Two  secants  meet  at  an  angle  of  35°;  the  nearer  of  the  inter- 
cepted arcs  is  50°.     What  is  the  value  of  the  farther  arc  ? 
Help.— 35°  =  \(x  -  50°). 


ADDITIONAL  THEOREMS 

302.  (1)  The  arcs  intercepted  on  a  circle 
by  two  parallel  chords  are  equal. 

(2)  The  arcs  intercepted  on  a  circle  by 
a  tangent  and  a  chord  parallel  to  it  are 
equal. 

(3)  The  bisector  of  an  inscribed  angle 
bisects  the  subtended  (or  intercepted)  arc. 

(4)  The   bisectors   of   all    the  angles 
inscribed  in   the  same  segment  are  con- 
current. 

State  where  the  bisectors  are  concurrent. 

(5)  Opposite    angles    of   an   inscribed  quadrilateral  are 
supplementary. 


CIKCLES 


203 


Help. — Consider  the  arcs  which  measure  these  angles. 

(6)  Corollary  of  Theorem  (5).     The  sum  of  one  pair  of 
opposite  angles  of  an  inscribed  quadrilateral  equals  the  sum 
of  the  other  pair  of  opposite  angles. 

(7)  The  value  of  angle  x  can  be  expressed  in  terms  of  arcs 
b  and  c.     Investigate,  and  state  the  theorem  discovered. 


(8)  The  value  of  angle  x  can  be  expressed  in  terms  of  arcs 
b}  c  and  d. 


Help.— Show  that  x  =  b  +  }(c  +  d). 
(9)  An  exterior  angle  of  an  inscribed  quadrilateral  equals 
the  interior  angle  at  the  opposite  vertex. 

(10)  An  angle  is  inscribed  in  each  of  the 
four    outside  segments  of  a  circle,  which 
are    formed  by  the  sides  of  an  inscribed 
quadrilateral.     Investigate  the  measure  of 
the  sum  of  these  four  angles. 

(11)  State  a  general  theorem  concerning 

the  measure  of  an  angle  in  terms  of  subtended  arcs,  accord- 
ing as  the  vertex  is  within,  on,  or  without  the  circle. 

(12)  An  angle  formed  by  two  lines  passing  through  the 
extremities  of  a  diameter  is  acute,  right,  or  obtuse,  accord- 


204  PLANE    GEOMETRY 

ing  to  the  position  of  the  vertex.     Investigate  and  state  the 
theorem. 

(13)  The  center  of  the  circumscribed  circle  of  a  triangle 
lies  within  the  triangle,  on  a  side,  or  without  the  triangle, 
according  to  the  angle-shape  of  the  triangle.  Investigate 
and  state  the  theorem. 

303.  The  Fourth  Step  in  Classification. — Make  an  experi- 
mental test  of  the  principles  of  circles  which  have  been 
deduced  from  the  postulates  and  other  experimental 
principles. 

1.  The  angle  formed  by  a  tangent  and  a  chord  drawn  from 
the  point  of  tangency  (297). 


B 

Draw  the  figure  of  297  large  enough,  or  extend  the  angle 
lines  of  the  figure  as  shown,  so  that  angles  BAC  and  AOC 
can  be  measured  with  a  protractor.  Compare  these  measure- 
ments and  check  the  principle  of  297. 

2.  The  angle  formed  by  two  chords  which  intersect  within 
a  circle  (298). 

6 

<K 
-0 


Measure  angle  AEC,  and  angles  FOG  and  HOK.  Note 
that  the  latter  angles  measure  the  subtended  arcs  LM  and 
NP.  Compare  with  the  principle  of  298. 


CIRCLES 


205 


3.  The  angle  formed  by  two  secants  which  intersect  outside 
a  circle  (299). 

D 

,6 


Measure  /.ABC,  and  ^sDOE  and  FOG;  etc. 

4.  The  angle  formed  by  a  secant  and  a  tangent  which 
intersect  outside  a  circle  (300). 

5.  The  angle  formed  by  two  tangents  which  intersect  outside 
a  circle  (301). 

6.  The  principle  of  302,  Theorem  (5). 

7.  The  principle  of  302,  Theorem  (7). 

8.  The  principle  of  302,  Theorem  (8). 

9.  The  principle  of  302,  Theorem  (9). 
10.  The  principle  of  302,  Theorem  (10). 

304.  The  Third  Step  in  Classification.    An  Investigation 
of  Sect  Properties  of  Lines  Drawn  to  a  Circle. — The  sects 
intercepted  by  a  circumference  on  any  two  intersecting 
lines  which  intersect  or  meet  a  circle  in  two  points  or  which 
are  tangent  to  the  circle,  possess  some  relative  or  propor- 
tional values.     The  simplest  of  these  sect  properties  are 
stated  in  63,  Postulate  11,  and  in  292.     Other  sect  relations 
which  are  not  evident  from  a  constructed  figure,  will  now 
be  deduced  from  the  principles  of  similar  triangles. 

305.  Theorem  XV. — A  relation  exists  between  the  lengths 
of  the  segments  of  two  chords  which  intersect 

within  a  circle. 

Helps.— (1)    Draw    AC    and    BD.     (2)  Ai 
AAEC  ~  ABED;  because  /.A  =  /.D  and 
Z.C  =  Z£;  why?     (3)  State  the  proportion 
between  sides  AE}  CE}  BE  and  DE.  (4) 
Change  the  proportion  into  products. 


206  PLANE    GEOMETRY 

Draw  the  two  triangles  under  consideration  separately, 
and  overlay  the  equal  angles  with  colored  crayons.  Then 
overlay  the  homologous  sides  of  the  triangles  which  enter 
into  the  required  proportions,  with  colored  crayons. 

Corollaries. — (1)   The  segments  of  all  chords  of  a  pencil 
of  chords   intersecting    within  a  circle   are 
related. 

2.  A  line  drawn  perpendicular  to  the 
diameter  of  a  circle  is  a  mean  proportional 
between  the  segments  of  the  diameter.  (See 
also  260,  Theorem  (20).) 

EXERCISES 

1.  Two  chords  10"  and  12"  respectively,  intersect  at  a  point  which 
is  4"  from  the  end  of  the  10"  chord.     How  far  from  the  end  of 
the  12"  chord  is  the  point  of  intersection? 

2.  A  third  chord  of  the  circle  is  concurrent  with  the  two  chords  of 
Exercise  1,  and  intersects  them  at  a  point  which  is  6"  from  its 
nearer  end.     What  is  the  length  of  the  chord? 

3.  The  diameter  of  a  circle  is  400  feet  and  a  perpendicular  is  erected 
to  the  diameter  at  a  point  which  is  40  feet  from  one  end  of  the 
diameter.     Calculate  the  length  of  the  perpendicular  from  the 
diameter  to  its  point  of  intersection  with  the  circle. 

306.  Theorem  XVI. — A  relation  exists  between  the  lengths 
of  two  secants  which  intersect  outside  a  circle  and  the  external 
segments  of  the  secants.  ^ ^ 

Helps. — The  overlapping  triangles 
are  similar.  State  the  proportions 
in  the  form  of  equal  products.  Fol- 
low the  suggestions  given  in  305. 

Corollary. — The  lengths  of  all  secants 
of  a  pencil  of  secants  intersecting  outside  of  a  circle,  and  of 
their  external  segments,  are  related. 

EXERCISES 

1.  Two  secants  are  20"  and  24"  measured  from  their  point  of  inter- 
section to  the  farther  point  of  intersection  with  the  circle.  The 
20"  secant  intersects  the  circle  12"  from  its  point  of  intersection 
with  the  other  secant.  At  what  distance  from  their  point  of 
intersection  does  the  24"  secant  cut  the  circle? 


CIRCLES 


207 


2.  The  secants  of  Exercise  1  meet  a  third  secant  at  the  same  point. 
A  length  of  1"  is  intercepted  on  this  third  secant  between  the 
two  points  of  intersection  with  the  circle.     What  is  the  total 
length  of  this  secant? 
Help.—x(x  +  1)  =  240. 

307.  Theorem  XVII. — A  relation  exists  between  the  lengths 

of  a  secant  and  a  tangent  drawn  to  a  /- \ 

circle  from  the  same  point  outside  a  A          B  /  \  C 

circle,  and  the  external  segment  of 

the  secant. 

Helps.  — AABD  ~  AACD. 
State  the  required  relation  as  a 
proportion.  Follow  the  suggestions  given  in  305. 

Corollary. — The  square  of  the  tangent  equals,  etc. 

EXERCISES 

1.  In  the  figure  of  Theorem  XVII,  the  segment  AB  =  10",  and 
segment  BC  —  12".     Calculate  the  length  of  the  tangent  AD. 

2.  The  length  of  a  tangent  AD  =  12",  and  the  length  of  the  secant 
AC  =  20".     Calculate  AB  and  BC. 

308.  THEOREM  XVIII. — The  radius  of  the  circumscribed 
circle  of  a  triangle  can  be  expressed  in  terms  of  the  three  sides 
of  the  triangle. 

A 


State  the  hypothesis  and  conclusion. 

Helps.— (1)  Draw  AE,  OD,  CD;  (2)  AABE  ~  AACD; 


(5)  but  pa  =  - 

(6)  .'.  R  = 


-a)(s-b)(s  -c);(255); 
abc 


—  a)(s  —  b)(s—  c) 


208  PLANE    GEOMETRY 

EXERCISES 

1.  Calculate  the  radius  of  the  circumscribed  circle  of  a  triangle  of 
which  the  sides  are  10,  12  and  20. 

2.  Calculate  the  radius  of  the  circumscribed  circle  of  a  triangle  of 
which  the  sides  are  20,  20  and  10. 

309.  THEOREM  XIX. — The  square  of  the  bisector  of  an 
angle  of  a  triangle  equals  the  product  of  the  sides  including  the 
angle;  less  the  product  of  the  segments  into  which  the  third 
side  of  the  triangle  is  divided  by  the  bisector. 

State  the  hypothesis  and  conclusion. 

A 


Helps.  —  (1)    Draw    the    circumscribed    circle;    produce 
AD  to  E,  dfaw  CE;  (2)  AADB  ~  AACE;  (3) 


(4)  xy  =  bA'DE.  In  order  to  eliminate  AE  and  DE;  proceed 
as  follows;  (5)  AE  =  bA  +  DE;  (6)  Substitute   from  (5) 

in  (3);   b    +  DE  =  y;  (7)  Solve  for  DE  in  (6)  and  sub- 
stitute in  (4);  ^=  b°  ~  bl  ;    (8)  .'.  bl  =  be  -  xy. 

OA  OA 

EXERCISES 

1.  The  sides  of  a  triangle  are  10,  12  and  20.     Calculate  the  length 
of  the  bisector  of  the  angle  opposite  the  12  side. 

Help.  —  (1)  Calculate  the  segments  x  and  y  by  246;  (2)  calculate 
bA  by  the  above  formula. 

2.  Calculate  by  the  above  formula;  the  length  of  the  bisector  of  the 
vertex  angle  of  an  isosceles  triangle  of  which  the  sides  are  20, 
20  and  10. 

310.  THEOREM  XX.  —  The  length  of  the  bisector  of  an  angle 
of  a  triangle  can  be  expressed  in  terms  of  the  three  sides  of 
the  triangle. 


CIRCLES  209 

Helps.—  (1)  b\  =  be  -  xy;   (2)    x  =    g^-j  ,y  =  j-~> 

(260,  Theorem  (26);  (3)  substitute  from  (2)  in  (1); 

...    .  ,, 
;  (4)  .Vti  - 


-c)2-a2-|       r(fl  +  6  +  C)(- 

(6+c)2    J=  (6  +  c) 


EXERCISES 

1.  The  sides  of  a  triangle  are  10,  12  and  20.     Calculate  the  length 
of  the  bisector  of  the  angle  opposite  the  12  side,  using  the  above 
formula. 

2.  Calculate  by  this  formula,  the  three  angle  bisectors  of  a  triangle 
of  which  the  sides  are  20,  20  and  10. 

311.  The  Fourth  Step  in  Classification. 

1.  Two  chords  which  intersect  within  a  circle  (305). 
Measure  the  chord  segments  to  the  nearest  hundredth 

of  an  inch.     Test  the  proportion  or  the  relation  of  products, 
as  discovered. 

2.  The  principle  of  305,  Corollary  2. 

3.  Two  secants  which  intersect  outside  a  circle  (306). 
Measure  the  secants  and  their  external  segments.     Test 

the  relation  discovered. 

4.  A  secant  and  a  tangent  which  intersect  outside  a  circle 
(307). 

5.  The  radius  of  the  circumscribed  circle  of  a  triangle  (308). 
Draw  any  triangle  and  its  circumscribed  circle.     Measure 

the  sides  of  the  triangle  and  the  radius  of  the  circle.     Test 
the  relation  discovered. 

6.  Test  the  relation  of  309  by  measurement. 

7.  Test  the  relation  of  310  by  measurement. 

312.  The  Position  Assigned  a  Principle  in  Classification. 
Notice    in   what   widely   separated   positions   apparently 
closely  related  principles  must  be  placed,  if  they  are  asso- 
ciated with  the  group  of  principles  by  which  they  are 

14 


210  PLANE    GEOMETRY 

analyzed  or  explained.     Thus  the  following  special  lines  of 
a  triangle  have  been  expressed  in  terms  of  the  three  sides : 

1.  Under  similar  triangles;  an  altitude  (255)  and  a  median 
(256). 

2.  Under  circles;  an  angle-bisector  (309  and  310)  and 
the  radius  of  the  circumscribed  circle  (308). 

3.  Under  areas;  the  radius  of  the  inscribed  circle,  which 
will  be  derived  in  Chapter  XVII. 

ADDITIONAL  THEOREMS 

313.  (1)  Analyze  Theorem  VIII  (297)  by  drawing  radii 
perpendicular  to  the  chord  and  to  the  tangent.  The  measure 
of  the  required  angle  can  be  obtained  from  that  of  the  cen- 
tral angle  thus  formed. 

(2)  Analyze  Theorem  VIII  (297)  by  drawing  a  chord 
through  the  farther  extremity  of  the  given  chord  parallel 
to  the  tangent.     Apply  302,  Theorem  (2). 

(3)  Analyze  Theorem  IX  (298)  by  drawing  an  auxiliary 
chord  through  the  extremity  of  one  of  the  given  chords 
parallel  to  the  other  given  chord.     Apply  302,  Theorem  (2) . 

(4)  Analyze  Theorem  X  (299)  by  the  method  of  Theo- 
rems (2)  and  (3). 

(5)  Analyze  Theorem  XI  (300)  by  the  method   of  the 
preceding  theorems. 

(6)  Analyze  Theorem  XII  (301)  by  the  method  of  the 
preceding  theorems. 

(7)  Analyze  Theorem  XI  (300)  by  considering  a  tangent 
as  a  special  case  of  a  secant  whose  points  of  intersection 
coincide,  thus  deriving  Theorem  XI  from  Theorem  X. 

(8)  Analyze  Theorem  II  of  Chapter  V  (119)  by  circum- 
scribing a  circle  about  the  triangle. 

(9)  Show  that  the  analysis  of  Theorem  8  is  faulty  logic 
because  the  theorem  itself  enters  into  the  analysis  through 
intermediate  theorems. 

(10)  A  chord  subtending  an  arc  of  120°  bisects  a  radius 
perpendicular  to  it. 

(11)  Tangents  drawn  from  an  exterior  point  a  radius  dis- 
tant from  the  circumference,  form  an  equilateral  triangle  with 
their  chord  of  contact. 


CIRCLES 


211 


(12)  Lines  drawn  from  the  vertices  of  a    circumscribed 
quadrilateral  to  the  center  of  the  circle,  form  central  angles  of 
which  each  non-adjacent  pair  are  supplementary. 

(13)  The  angle  between  two  tangents  to  a  circle  is  double  the 
angle  between  the  chord  of  contact  and  a  radius  drawn  to  a 
point  of  tangency. 

(14)  A  circumference  whose  diameter  is  the  hypotenuse  of 
a  right  triangle,  contains  the  vertex  of  the  right  angle. 

(15)  A  circle  drawn  with  one  of  the  equal  sides  of  an  isos- 
celes triangle  as  a  diameter,  bisects  the  base  of  the  triangle. 


Helps.— Draw  AD;  AD±BC,  why?  .*.  AD  bisects  BC. 

(16)  A  line  is  drawn  through  the  point  of  contact  of  two 
tangent  circles  (externally  or  internally  tangent).  Radii 
drawn  to  the  extremities  of  this  line  are  parallel. 


Helps. — Draw  the  line  of  centers.  Show  that  alternate- 
interior  angles  are  equal. 

(17)  A  line  is  drawn  through  the  point  of  contact  of  two 
tangent  circles  (two  cases).     Tangents  drawn  at  the  extremities 
of  this  line  are  parallel. 

Helps. — (a)  Draw  radii  as  in  Theorem  (16);  or  (6)  Draw 
a  common  tangent,  and  show  that  alternate-interior  angles 
are  equal  because  their  arcs  have  the  same  measure. 

(18)  The  perpendicular  from  the  center  of  a  circle  to  aside 


212 


PLANE    GEOMETKY 


of  an  inscribed  equilateral  triangle  equals  one-half  the  radius 
of  the  circle. 

(19)  The  internal  common  tangent  of  two  externally  tangent 
circles  bisects  the  external  common  tangents. 


(20)  Tangents  drawn  to  two  tangent  circles  from  a  point  on 
their  internal  common  tangent  are  equal  (two  cases). 


(21)  A  circle  is  drawn  with  the  radius  of  another  circle  as 
its  diameter.    A  chord  of  the  larger  circle  drawn  from  the 
point  of  tangency  of  the  two  circles  is  bisected  by  the  circum- 
ference of  the  smaller  circle. 

(22)  All  chords  drawn  from  the  point  of  tangency  of  two 
circles  (internally  or  externally  tangent)  are  divided  propor- 
tionally by  the  circumference  of  the  smaller  circle. 

6 


Helps. — Draw  a  common  diameter;  AABF 
AABD  ~  AACE,  etc. 


AACGand 


CIRCLES 


213 


(23)  Two  (unequal)  circles  intersect  in  points  A  and  B; 
a  common  secant  is  drawn  through  A  intersecting  the  circles 
in  C  and  D;  chords  BC  and  BD  are  drawn.     Then   ^CBD 
is  the  same  for  any  position  of  the  secant  CD;  or,    /.CBD  is 
constant. 

Helps. — Consider  the  measures  of  angles  C  and  D  and  how 
Z.CBD  is  related  to  these  angles. 

(24)  The  circles  described  on  two  sides  of  a  triangle  intersect 
on  the  third  side. 


Helps. — Draw  AD.LBC;  show  that  each  circle  contains 
point  D. 

(25)  Of  all  triangles  drawn  with  the  same  base  and  with 
equal  altitudes,  the  isosceles  triangle  has  the  largest  vertex 
angle. 


Helps. — Circumscribe  a  circle  about  the  isosceles  triangle 
ABC,  what  kind  of  line  is  AA'A"?  .'.  points  A',  A",  etc., 
are  outside  the  circle. 

(26)  A  chord  is  a  mean  proportional  between  a  diameter 
drawn  through  one  of  its  extremities  and  its  projection  on  the 
diameter. 

(27)  The  sum  of  the  three  alternate  angles  of  an  inscribed 
hexagon  equals  the  sum  of  the  other  three  alternate  angles. 

(28)  Find  in  a  way  similar  to  that  employed  in  the  pre- 
ceding theorem,  that  a  relation  exists  between  the  sums  of 
the  two  sets  of  alternate  angles  of  any  inscribed  polygon 
of  an  even  number  of  sides.     State  the  theorem. 


214 


PLANE    GEOMETRY 


(29)  The  point  of  intersection  of  any  pair  of  common 
tangents  of  two  circles  divides  the  center  sect  into  segments 
proportional  to  the  radii. 

(30)  Corollary  of  (29).     The  points  of  intersection  of  the 
internal  and  external  common  tangents  of  two  circles  divide 
the  center  sect  harmonically. 

(31)  A  pencil  of  chords,  AB,  AC,  AD,  etc.,  is  drawn  from 
a  point  A  in  the  circumference  of  a  circle;  a  line  parallel  to 
the  tangent  at  A,  intersects  the  chords  in  points  B',  C',  Df, 
etc.     Then  AB-AB'  =  AC-AC'  =  etc. 

Helps. — Draw  a  diameter  AX  perpendicular  to  the  tan- 
gent at  A;  drsiwBX,  CX,  DX,  etc.;  &AB'X'~&ABX,  etc. 

(32)  The  common  chord  of  two  intersecting  circles,  if  pro- 
duced bisects  their  common  tangents. 

(33)  If  the  altitudes  of  a  triangle  are  produced  to  intersect 
the   circumscribed  circle,   the   sects  intercepted  between   the 
orthocenter  and  the  circumference  are  bisected  by  the  sides  of 
the  triangle. 


Helps.— (1)  x  =  y;   (2)  x  =  z;  (3)  :.  y  =  z ;  (4)  hence 
OG  =  GD. 

(34)  An  internal  common  tangent  of  two  equal    circles, 
which  are  externally  out  of  contact,  bisects  the  center  sect. 

(35)  Secants  drawn  from  the  point  of  intersection  of  the 
internal  and  external   common   tangents   of  two   externally 
tangent  circles,  to  the  centers  of  the  circles,  are  perpendicular. 


CIRCLES  215 

(36)  The  perpendicular  bisectors  of  the  sides  of  an  inscribed 
polygon  are  concurrent  at  the  center  of  the  circle. 

(37)  The  bisectors  of  the  angles  of  a  circumscribed  polygon 
are  concurrent  at  the  center  of  the  circle. 

(38)  The  perimeter  of  a  circumscribed  trapezoid  equals 
four  times  the  sect  which  joins  the  middle  points  of  the  two 
non-parallel  sides. 


(39)  The  sum  of  three  alternate  sides  of  a  circumscribed 
hexagon  equals  the  sum  of  the  other  three  alternate  sides. 

(40)  Test  the  principles  of  Theorems  (38)  and  (39)  ex- 
perimentally. 

314.  Locus  CONSTRUCTIONS. — Construct  the  following 
loci  experimentally,  by  finding  the  positions  of  a  sufficient 
number  of  points  to  indicate  the  form  of  the  complete  locus. 

1.  The  locus  of  the  middle  points  of  a  system  of  parallel 
chords  of  a  circle. 

2.  The  locus  of  the  middle  points  of  all  equal  chords  of  a 
circle. 

3.  The  locus  of  the  middle  points  of  all  chords  of  a  circle 
which  are  concurrent  in  a  point  on  the  circumference. 

4.  The  locus  of  middle  points  of  all  chords  of  a  circle 
which  are  concurrent  in  a  point  within  the  circle. 

5.  The  locus  of  the  middle  points  of  all  secants  of  a 
circle  which  are  concurrent  in  a  point  outside  the  circle. 
Plot  the  bisection  points  both  of  the  entire  secants  and  of 
their  external  segments. 

6.  The  locus  of  the  centers  of  all  circles  which  are  tangent 
to  a  given  straight  line  at  a  given  point. 

7.  The  locus  of  the  centers  of  all  circles  tangent  to  a 
given'  circle  at  a  given  point.     Draw  both  externally  and 
internally  tangent  circles. 

8.  From  a  given  point  on  the  given  straight  line  of 


216  PLANE    GEOMETRY 

Exercise  6,  tangents  are  drawn  to  all  the  circles.     Find 
the  locus  of  the  points  of  tangency. 

9.  The  locus  of  the  vertices  of  all  angles  of  a  given  size, 
or  value,  whose  sides  pass  through  two  given  points.     Mark 
two  points  several  inches  apart,  and  use  the  30°  angle  of  a 
drawing  triangle  for  the  given  angle.     In  order  to  complete 
the  locus  it  is  necessary  to  find  positions  of  the  angle  where 
one  side  produced  through  the  vertex,  contains  one  of  the 
given  points. 

315.  REVIEW  EXERCISES 

1.  A  secant  and  a  tangent  intersect  at  an  angle  of  30°;  the  secant 
intercepts  an  arc  of  100°  between  its  points  of  intersection  with 
the  circle.     Calculate  the  other  two  intercepted  arcs. 

2.  Two  angles  of  an  inscribed  triangle  are  48°  and  105°;  calculate 
the  arcs  subtended  by  the  sides. 

3.  The  vertex  angle  of  an  inscribed  isosceles  triangle  is  50°;  calcu- 
late the  arcs  subtended  by  the  sides. 

4.  Two  secants  intercept  arcs  of  150°  and  95°;  calculate  the  angle 
between  them. 

5.  Two  secants  form  an  angle  of  30°,  the  nearer  intercepted  arc 
being  50°;  calculate  the  farther  of  the  intercepted  arcs. 

6.  The  measure  of  the  farther  arc  intercepted  by  two  secants  is 
four  times  the  measure  of  the  angle  formed  by  the  secants;  the 
nearer  arc  is  30°;  calculate  the  angle  and  remoter  arc. 

7.  Two  tangents  form  an  angle  of  25°;  calculate  the  major  and 
minor  intercepted  arcs. 

8.  An  inscribed  angle  subtends  an  arc  of  75°;  calculate  its  value. 
What  arc  is  subtended  by  an  inscribed  angle  of  75°? 

9.  Two  intersecting  chords  intercept,  in  order,  the  arcs  80°,  120°, 
30°,  130°;  calculate  the  angles  formed  by  the  chords. 

10.  One  of  the  angles  formed  by  two  intersecting  chords  is  55°;  an 
arc  opposite  this  angle  is  78°,  and  an  adjacent  arc  is_100°. 
Calculate  the  other  arcs. 

11.  A  chord  is  drawn  from  a  point  of  tangency  making  an  angle  of 
42°  with  the  tangent;  calculate  the  values  of  the  two  subtended 
arcs. 

12.  A  chord  subtends  an  arc  of  75°;  at  what  angle  must  a  line  inter- 
secting it  at  an  extremity  be  drawn  in  order  to  be  a  tangent  to 
the  circle? 

13.  Three  consecutive  sides  of  an  inscribed  quadrilateral  subtend 

90°,  58°,  calculate  the  angles  of  the  quadrilateral, 


CIRCLES  217 

the  angles  between  the  diagonals,  and  the  angles  at  which  the 
diagonals  meet  the  sides. 

14.  Two  tangents  are  drawn  at  the  extremities  of  a  chord  which 
subtends  an  arc  of  155°;  at  what  angle  do  the  tangents  intersect? 

15.  Two  perpendicular  chords  intercept  adjacent  arcs  of  40°  and 
70°;  calculate  the  other  intercepted  arcs. 

16.  Two    adjacent   arcs   AB   and   BC,   measure  150°  and  100°; 
a  tangent  is  drawn  at  B,  and  a  secant  through  AC;  calculate 
the  angle  of  intersection. 

17.  A  chord  2"  long  is  drawn  perpendicular  to  a  diameter  of  a  circle 
of  1.5"  radius;  at  what  point  does  the  chord  intersect  the 
diameter? 

18.  A  chord  of  1"  is  bisected  by  a  chord  on  which  a  segment  of 
\"  is  cut  off;  calculate  the  radius  of  the  circle. 

19.  From  the  end  of  a  tangent  sect  of  2",  a  secant  is  drawn  through 
the  center  of  a  circle  of  3"  diameter;  calculate  the  external  seg- 
ment of  the  secant. 

20.  The  sides  of  an  isosceles  triangle  are  4,  4,  3;  calculate  the  three 
angle-bisectors,  and  the  radius  of  the  circumscribed  circle. 

21.  Calculate  the  radius  of  the  circumscribed  circle  of  a  triangle 
whose  sides  are  3,  4,  5. 

22.  The  radius  of  a  circle  is  2";  calculate  the  chord  subtending  an 
arc  of  120°. 

23.  Two  circles  of  8"  and  10"  radii  have  a  center  sect  of  20";  calcu- 
late the  tangent  sects  of  internal  and  external  common  tangents. 

24.  In  the  formulas  of  308  and  310,  let  a  =  b  =  c.     Show  that 

R  =  ;  6  A  =  —~ — ;  and  that  these  values  agree  with  values 

3  '         2 

for  the  altitude  and  median  of  an  equilateral  triangle,  derived 
in  a  similar  way  from  the  formulas  for  pa  and  ma,  255  and  256 
(see  260,  Theorem  (16);  and  with  254,  Part  (6). 

25.  Given  an  angle,  and  the  length  of  a  chord;  construct  a  circle 
such  that  the  segment  cut  off  by  the  chord  contains  inscribed 
angles  equal  to  the  given  angle. 

26.  Given  a  chord  of  a  circle  and  the  measure  of  its  subtended  arc; 
construct  the  circle. 

27.  Divide  a  circle  into  two  segments  such  that  the  angle  inscribed 
in  one  segment  is  double  the  angle  inscribed  in  the  other. 

28.  Construct  a  circle  tangent  to  each  of  two  parallel  lines  and  to  a 
third  line  intersecting  the  parallels. 

29.  Construct  a  circle  tangent  to  each  of  two  concentric  circles. 
Write  a  formula  for  its  radius. 

30.  Construct  a  circle  tangent  to  each  of  two  concentric  circles 
and  also  tangent  to  a  line  intersecting  one  or  both  of  the  given 
circles. 


218  PLANE    GEOMETRY 

31.  Construct  a  circle  of  given  radius,  tangent  to  each  of  two  given 
circles  which  are  externally  out  of  contact;  discuss  all  cases. 

32.  The  sides  of  a  triangle  are  5",  6",  8".     Calculate  the  segments 
into  which  the  sides  are  divided  by  the  points  of  tangency  of 
the  inscribed  circle. 


Helps.  —  AF  =  AE,  etc.     Form  a  system  of  equations,  x  -f  y  =  5, 
etc.;  solve  for  x,  y,  z. 

33.  Derive  formulas  for  the  segments  into  which  the  sides  a,  6,  c, 
of  a  triangle  are  divided  by  the  points  of  contact  of  the  inscribed 
circle. 

34.  Calculate  the  distance  of  the  center  of  the  inscribed  circle  of 
the  triangle  of  Exercise  32  from  each  vertex  of  the  triangle. 
Helps.  —  Calculate  r;  then  d,A,  etc. 

35.  Show  that  the  general  formula  for  d,A  is, 


=     l 


-  a)  +  (s  -  b)(s  - 


36.  Three  circles  are  drawn  externally  tangent,  each  to  the  other 
two.  The  center  sects  are  10,  12  and  15  inches.  Find  the 
radii  of  the  circles. 


37.  Two  circles  are  externally  tangent  at  A.     One  of  the  external 
common  tangents  touches  the  circles  at  points  B  and  C,  respec- 
tively.    Show  that  a  circle  whose  diameter  is  BC,  contains  point 
A. 

38.  Show  that  two  equal  externally  tangent  circles  intercept  equal 
chords  on  a  common  secant  drawn  through  the  point  of  tangency. 


CIRCLES 


219 


39.  The  triangle  ABC  is  formed  by  two  fixed  tangents  AD  and  AE 
and  a  movable  tangent  BC.  Show  that  the  perimeter  of  AABC 
is  constant,  wherever  tangent  BC  is  drawn. 


40.  The  altitude  to  the  hypotenuse  of  a  right  triangle  divides  the 
hypotenuse  into  segments  of  10  and  40  inches  respectively. 
Calculate  the  altitude. 

41.  Show  that,  if  the  perpendicular  sides  of  a  right  triangle  are  in 
the  ratio  of  1  to  2,  the  altitude  to  the  hypotenuse  divides  it  in 
the  ratio  of  1  to  4. 

42.  A  surveyor  measured  a  chord  AB  of  a  circular  railroad  curve, 
equal  to  240  feet,  and  the  perpendicular  distance  CD  at  the 
middle  of  chord  AB,  equal  40  feet.     Show  how  to  calculate  the 
radius  of  the  curve. 


43.  The  altitude  to  the  hypotenuse  of  a  right  triangle  of  which 
one  acute  angle  is  30°,  divides  the  hypotenuse  in  the  ratio 
of  1  to  3. 

44.  Two  chords  of  6  and  10  inches  respectively,  intersect  so  as  to 
divide  the  shorter  chord  into  segments  of  2  and  4  inches.     Find 
the  segments  of  the  longer  chord. 

45.  A  secant  and  a  tangent  are  drawn  from  the  same  external  point. 
The  chord  intercepted  on  the  secant  is  8  inches  and  the  external 
segment  of  the  secant  is  14  inches.     Find  the  length  of  the 
tangent. 

46.  The  radius  of  a  circle  is  21  inches.     Find  the  length  of  a  tangent 
drawn  from  a  point  35  inches  from  the  center  of  the  circle. 
Find  also  the  length  of  a  secant  drawn  from  the  same  point  if 
the  length  of  its  external  segment  is  20  inches. 

47.  Tangents  drawn  to  two  intersecting  circles  from  any  point  on 
their  common  chord  produced  are  equal. 


220  PLANE    GEOMETRY 

48.  The  common  chords  of  each  pair  of  three  intersecting  circles 
are  concurrent. 


Helps. — (1)  AB  and  CD  intersect  in  0;  draw  chord  EO  which 
intersects  circle's  x  and  z  in  F  and  G,  respectively;  (2) A  O'OB 
=  EO-OF,  etc.;  (3)  .*.  OF  =  OG,  and  therefore  points  G  and  F 
coincide. 

49.  The  center  line  of  two  circles,  which  are  externally  out  of  con- 
tact, intersects  the  circles  in  the  successive  points  A,  B,  C,  D, 
and  intersects  their  common  external  tangent  in  point  E. 
Show  that  EA'ED  =  EB'EC. 

Helps. — (1)  Mark  the  points  of  tangency,  T  and  T';  draw  chords 

jf>  A        FC1 
AT,  BT,  CT',  DT';  (2)  J^  =  |^,,  etc. 

50.  In  Exercise  49,  draw  a  secant  from  point  E  intersecting  the 
circles  in  points  F,  G,  H,  J.     Show  that  EF-EJ  =  EG- EH. 

51.  The  radii  of  two  eircles  are  10  and  4  inches,  respectively,  and 
their  center  sect  is  20  inches.     Find  the  points  of  intersection 
of  the  internal  and  external  common  tangents  with  the  center 
sect. 

4  x  ,4  y 

To  =  20=~x and  10  - 20+7 

52.  The  radii  of  two  concentric  circles  are  10  and  24  inches  respec- 
tively.    Find  the  length  of  a  chord  of  the  larger  circle  that  is 
tangent  to  the  smaller  circle. 

53.  The  radius  of  a  circle  is  24  inches.     Find  the  length  of  the 
shortest  chord  that  can  be  drawn  through  a  point  20  inches 
from  the  center  of  the  circle. 

APPLICATIONS 

316.  FAMOUS  PROBLEMS. — The  Trisection  of  an  Angle — 
This  is  one  of  the  famous  problems  of  antiquity.  There  is 
no  known  solution  by  use  of  rule  and  compass  alone.  The 
following  method  requires  the  use  of  a  scale,  and  would  not 
be  considered  a  solution  by  those  old  Greek  geometers  who 


CIRCLES 


221 


regarded   the   science   chiefly   as   a   splendid   intellectual 
recreation.  C 


/.ABC  is  the  given  angle.  Draw  any  circle  with  center 
B;  produce  AB;  find  by  trial,  using  a  scale,  a  sect  ED  such 
that  EF  =  BF.  Then  x  =  %/.ABC,  and  can  be  laid  off 
at  ABG,  etc. 

EXERCISES 

1.  Prove  that  /x  =  I/ABC. 

2.  A  method  of  trisecting  an  angle,  which  has  been  frequently 
suggested  by  students  is  as  follows :  Draw  BD 

=  BE;  draw  DE;  trisect  DE  by  some  purely 
"geometrical"  method,  as  in  261,  Exercise  31; 
draw  BF  and  BG.  Prove,  that  /x  does  not 
equal  /y  by  showing  that  the  bisector  of  /.DBG 
does  not  bisect  DG,  245.  D  A 

3.  Can  the  error  in  the  construction  of  Exercise  2  be  detected  by 
measurement? 

The  Division  of  a  Sect  in  Mean  and  Extreme  Ratio. — This 
is  such  a  division  of  the  sect  that  one  segment  is  a  mean 
proportional  between  the  given  sect  and  the  other  segment. 
There  are  two  points  of  division,  one  internal  and  the  other 
external. 


H 


C  E  B 

—    x. >|«-,y-x — W 

—     5 H 


AB  is  the  given  sect.     Draw  circle  D,  tangent  to  AB  at 
,  with  radius  =  \  AB;  draw  AD.     Then  E  is  the  required 


222 


PLANE    GEOMETRY 


internal  point  of  division,  if  AE  =  AF;  and  H  is  the  required 
external  point  if  AH  =  AG. 


EXERCISES 


4.  Prove 


s  —  x 


Helps. — (1)  -£g  =  -£Q;  (2)  change  by  the  Law  of  Division  (240). 

5.  For  the  external  point  of  division  H,  the  segments  of  sect  AB 
sue  HA  =  s  +  x  and  HB  =  2s  +  x.     Prove  2s  +  x  =  lit? 

S  +  X  S 

6.  Another  construction  for  the  division  of  a  sect  into  mean  and 

s  —  x      x 


extreme  ratio  is  shown.  Prove 


x 


7.  Calculate  the  values  of  the  segments  of  a  sect  divided  in  mean 
and  extreme  ratio.     Solve  the  equation  of  Exercises  4  and  6  by 

-i  ±V8 


algebra.     Show  that  x 

& 

8.  In   Fig.    1,   let  HA  =  xir-  then 


•s   =  +.618s  or  -1.618s. 
=  — •      Solve  for    x\; 


'  s.      Show  that  x\  =  s  +  x  or  —  x,  and  there- 

et 

fore  gives  the  points  of  division  H  and  E,  which  were  found  in 
Exercise  7. 

9.  Measure  the  segments  of  the  sects  AB  of  the  preceding  figures. 
Do  the  measurements  agree  with  the  calculated  values? 

Artistic  Value  of  Mean  and  Extreme  Ratio. — This  division 
of  a  sect  possesses  a  remarkable  artistic  and  psychological 


CIRCLES 


223 


interest.     It  has  been  called  the  " divine  ratio"  or  the 
"  golden  section." 

EXERCISES 

10.  Draw  several  equal  vertical  lines,  side  by  side,  and  divide  them 
at  different  points.     It  will  be  found  that  many  persons  possess  a 
marked  preference  for  the  "divine  ratio,"  in  which  the  point 
of  division  is  above  the  center;  that  is,  x  =  .62s. 

11.  Measure  the  most  pleasing  proportioned  book  covers,  platters, 
"oval"  (elliptical)  picture  frames,  etc.     It  will  be  found  that 
the  width  approximates  .62  of  the  length. 

A  Mean  Proportional. 


EXERCISES 

12.  Explain  the  construction  of  the  figure  and  prove  that  x  is  the 
required  sect. 

13.  Calculate  the  value  of  x  and  check  the  geometrical  solution  by 
measurement. 

The  Geometrical  Solution  of  a  Quadratic  Equation.— 
C 


B 


Let  the  given  equation  be  x2  —  5x  +  4  =0.     Then  a;  (5  — 

x         2 

x)  =  4; .'.  ~  =  ^ —  -  •    Draw  a  circle  with  AB  =  5  units  as 
z     o  —  x 

diameter;  draw  AC JLA5,  equal  2;   draw   CD\\AB,   draw 
DE±AB.     Then  A#  and  EJ5  are  the  roots  of  the  equation. 

EXERCISES 

14.  Solve  the  equation  x2  —  9x  +  14  =0,  geometrically. 

15.  Solve  the  equation  xz  —  Wx  +  25  =  0,  geometrically. 

16.  Solve  the  equation  x2  —  6x  =  0,  geometrically. 

17.  Check  the  geometrical  solutions  by  algebraic  solutions  of  the 
preceding  equations. 


224 


PLANE    GEOMETRY 


317.  Terrestrial  Measurements. — The  curvature  of  a 
sphere  in  a  distance  d,  is  expressed  in  terms  of  the  intercept 
c,  on  a  radius,  included  between  the  arc  and  the  tangent. 

EXERCISES 

1.  In  the  right  triangle  BAG,  OB2  =  GA2_   __  A       d      B 

+  ZB2.     Substitute  the  lengths  r,  c^ 
and  d;  omit  the  term  c2  which  will  be 
very  small  in  comparison  with  the  other 

terms;  solve  for  c.     Show  that  c  =  »-• 

Solve  also  for  r  and  d.  In  these  form- 
ulas, d'  may  be  substituted  for  d,  when 
d  is  small  in  comparison  with  r. 

2.  Let  r  =  100  inches  and  d  =  20  inches. 
Find  c. 

3.  Let  d  =  20  inches  and  c  =  1  inch.     Find  r. 

4.  Let  r  =  120  inches  and  c  =  4  inches.     Find  d. 

The  Size  of  the  Earth. — Careful  measurements  prove  that 
the  curvature  of  the  earth  is  approximately  8  inches  in  1 
mile. 

EXERCISE 

5.  Find  the  radius  of  the  earth  from  the  preceding  data,  using  the 
formula  for  r. 

The  Distance  of  the  Horizon  at  Sea. — This  depends  upon 
the  height  of  the  observer  above  the  mean  surface  of  the 
water. 

EXERCISES 

6.  Find  the  distance  of  the  horizon  to  an  observer  who  stands  upon 
a  boardwalk  so  that  his  eye  is  30  feet  above  the  water  level  of 
the  ocean. 

30 
Helps. — c  =  £^7;  miles;  let  r  =  3960  miles,     d  =  \/2cr  miles. 


d' 


7.  Find  the  distance  at  which  the  topmast 
of  a  vessel  is  visible  to  the  observer  of 
Exercise  6,  if  the  topmast  is  80  feet 
above  the  water  line. 

80 
Helps. — For  the  vessel,  c'  =^280'    The 

distance  required  =  d  of  Exercise  6  + 
d'  of  Exercise  7. 


CIRCLES 


225 


8.  How  high  must  a  war  balloon  ascend  at  the  sea  coast  in  order  to 
sight  an  enemy  fleet  100  miles  distant? 

Help. — A  more  exact  result  will  be  obtained  in  this  problem  by 
solving  the  equation  (c  +  3960) 2  -  39602  =  1002. 

9.  How  far  out  at  sea  may  a  lighthouse  be  visible  if  the  light  is 
120  feet  above  water  level,  and  the  watch  on  the  vessel  is  60  feet 
above  the  water  level?     The  effective  distance  will  be  consider- 
ably less  than  that  calculated  except  under  most  favorable 
weather  conditions. 

318.  Navigation.     The  "Danger  Angle." — The  reefs  or 
bars  C,  D,  E,  etc.,  are  shown  on  a  sailing  chart,  A  and  B 


'       1 


\  /        \   -  \          / ' 

\  ,\'  ^}  //  J> 

r-..*£K  \  C&!   °' 


being  the  positions  of  two  prominent  objects  also  marked 
on  the  chart.  The  navigator  draws  circles  passing  through 
A  and  B  (or  they  are  already  shown  on  the  chart),  in  such 
a  way  as  to  include  or  exclude  the  danger  region.  He 
measures  any  inscribed  angles  x  and  y.  If  he  then  lays  his 
course  on  the  line  FG  so  that  the  sextant  reading  of  the  angle 
ASB  is  less  than  x  and  greater  than  y,  the  course  is  safe. 
Angles  x  and  y  are  the  "  danger  angles." 

EXERCISE 

1.  Explain  why  the  method  employed  ensures  a  safe  course. 

The  Three  Point  Problem. — This  name  is  given  to  a 
method  used  for  accurately  locating  a  ship  which  is  near 
enough  to  the  shore  to  distinguish  three  prominent  objects. 

15 


226 


PLANE    GEOMETRY 


The  angles  x  and  y  are  measured  with  a  sextant  on  the 
ship.     The  circles  0  and  0'  can  then  be  constructed,  their 


intersection  on  the  chart  determining  the  position  of  the 
ship  at  S. 

EXERCISES 

2.  What  relation  exists  between  angles  x  and  R;  T  and  yf 

3.  What  relation  exists  between  angles  z  and  R;  w  and  Tl 

4.  Prove  therefore  that  z  =  90°  —  x,  and  w  =  90°  —  y. 

5.  Plot  three  points  A,  B,  C;  AB  =  20,000  ft.,  BC  =  15,000  ft., 
Z.ABC   =   160°.     Find  the  position  of  the  ship  S  if  the  angle  x 
is  measured  40°,  and  angle  y  is  measured  35°. 

319.  Surveying.  Location  of  Soundings. — The  "  three 
point  problem'7  is  generally  used  to  locate  the  boat  which 
makes  the  soundings  along  a  shore  line,  or  within  a  lake  or 

0 


bay.  The  series  of  points  A,  B,  C,  D,  etc.,  are  located  and 
plotted  on  the  shore.  At  each  observation  point,  1,  2,  3, 
etc.,  angles  x,  y,  x',  y';  etc.,  are  measured  with  a  sextant, 
and  simultaneously  a  sounding  is  taken.  A  record  is  kept, 
from  which  the  soundings  can  be  plotted  on  the  map  or 
chart. 


CIRCLES 


227 


Point 

1st  angle 

2nd  angle 

Sounding,  feet 

1 

AB  75°  10' 

BC  42°  15' 

12.5 

2 

AB  50°    5' 

5C61°    3' 

24.3 

3 

4 

5 

BC  31°  52' 

CD  35°  37' 

27.8 

Etc. 

EXERCISE 

1.  Plot  the  points  1,  2  and  5,  from  the  record. 

To  Lay  Out  a  Semicircular  Curve  by  Ordinates.' 

0' 


QDEFG 

The  diameter  AB  is  divided  into  any  number  of  equal  parts, 
AD,  DE,  EF,  etc.,  with  a  smaller  part  AC  =  KB  at  the 
ends.  The  ordinates  CC',  DD',  etc.,  are  calculated  by  30  6 
Cor.  2. 

EXERCISES 

2.  If  AB  =  400  feet,  AC  =  CD  =  20  feet,  DE  =  EF  =  FG  =  etc. 
=  40  feet,  calculate  CC',  DD',  etc.,  to  OO'. 

3.  Plot  the  diameter  and  points  of  division,  draw  the  ordinates, 
measure  off  the  lengths  of  the  ordinates.     Test  the  accuracy  of 
the  calculated  lengths  by  drawing  a  circle  with  center  at  O. 

320.  Astronomy.  The  Height  of  the 
Moon's  Mountains. — Galileo,  about 
the  year  1600,  measured  the  heights 
of  some  of  the  mountains  on  the 
moon  by  the  following  method.  He 
measured  the  distance  AM  at  the 
instant  that  the  sun's  ray  SA,  first 
iluminated  the  top  of  the  mountain  M . 
the  moon  must  also  be  known. 


The  diameter  of 


228 


PLANE    GEOMETRY 


EXERCISE 

1.  If  AB  =  2160  miles,  and  AM  is  measured  120  miles,  calculate 
the  height  of  the  mountain  M . 

Help. — (a)   Use  the  formula  of  317,  Exercise   1;  and  (6)  use 
the  method  of  317,  Exercise  8. 

Eclipses  of  the  Sun  and  Moon. — The  eclipse  of  the  sun  is 
caused  by  the  converging  cone  of  total  shadow  of  the  moon 
falling  upon  the  earth's  surface  at  P  (Fig.  1).  The  eclipse 
of  the  moon  is  caused  by  the  entrance  of  the  moon  into  the 
cone  of  total  shadow  of  the  earth  (Fig.  2). 


FIG.  2. 

EXERCISES 

2.  Take  the  distance  from  center  to  center  of  the  sun  and  earth  as 
93,000,000  miles;  the  distance  from  the  earth  to  the  moon  as 
241,500  miles;  the  radius  of  the  sun  432,000  miles;  the  radius  of 
the  earth  3960  miles;  the  radius  of  the  moon  as  1080  miles.     (The 
distances  from  the  earth  to  the  sun  and  to  the  moon  are  variable.) 
Calculate  the  length  MF  of  the  moon's  shadow,  Fig.  1.     Does 
the  moon's  shadow  reach  the  surface  of  the  earth?     Calculate 
the  radius  of  the  shadow  at  P. 

SE  -  EB 

3.  In  Fig.  2,  show  that  EG  =  ^  _  jg  * 

4.  Calculate  the  length  of  the  earth's  total  shadow,  using  the  values 
given  in  Exercise  2  and  the  formula  of  Exercise  3. 


CIRCLES 


229 


5.  Calculate  the  diameter  of  the  earth's  shadow,  Fig.  2,  at  the  dis- 
tance at  which  the  moon  enters  it.  If  the  moon  enters  the 
shadow  centrally,  is  the  shadow  sufficiently  broad  to  entirely 
obscure  the  moon? 

321.  THE  CIRCLE  IN  DESIGN. — Ornamental  designs  in- 


FIG.  10. 

volving  circles  and  arcs  of  circles  are  found  in  railings, 
brackets,    cut   glass,    window   glass,   rugs   and   linoleums, 


230 


PLANE   GEOMETRY 


and  in  innumerable  other  familiar  objects.     A  few  sugges- 
tive illustrations  are  given. 

Figure  1  is  called  the  curved  equilateral  triangle.  Figures 
5  and  6  are  art  glass  designs;  Fig.  8  is  a  column  base;  Fig. 
10  shows  various  forms  of  mouldings  and  cornices. 

EXERCISES 

1.  Make  other  designs  for  some  definite  application,  as  a  piece  of 
scroll  work  to  fill  in  the  top  of  an  arched  doorway;  a  piece  of 
embroidery  or  a  sofa  cushion;  an  arm  of  a  built-in  bench  for  a 
porch  or  hall;  etc. 

2.  Name  some  familiar  objects  which  are  always  or  frequently 
circular  in  form. 

Architectural  Designs  Containing  Tangent  Circles. — These 
forms  occur  in  endless  variety.  Constructions  may  be 
made  by  trial,  but  interesting  applications  of  algebraic 
calculation  may  be  made,  and  should  be  made  in  laying 
out  exact  work.  Construct  the  figures  shown,  from  the 
calculated  values. 

EXERCISES 

3.  Express  the  radius  r,  of  one  of  the  circles  in  terms  of  the  dis- 
tance, d. 


Helps.— CD  =  r\/3;  d  =  CD  +  2r  =  r(2  +  \/3); 

=  d(2  -  V3)  =  -27d. 
4.  Express  r  in  terms  of  d. 


2+  x/3 


CIRCLES 


231 


Helps.— CD  =  rV3;  d  =  r(I  +  \/3); 
5.  Express  r  in  terms  of  d. 


Help.—  r  = 


d( V5  -  1) 


6.  Express  r  and  AB  in  terms  of  d. 


Helps.— In  &ABD,AB2  =  (d  -  r)2  - 
In  AABC,  AB2  =  (r  +  ^)  *  -  gV  = 


=  -r    -  2rfr  +  r2. 


and  A£  =  .49^. 
7.  Express  r  and  AB  in  terms  of  d. 


Helps.— r  =  -A 


232  PLANE   GEOMETRY 

8.  Express  r  and  A  B  in  terms  of  d. 


Helps.— r  =  ~  d  and  AB  =        d. 
9.  Express  r  and  BC  in  terms  of  d. 


Helps.— In  ACDE,  DE  = 

+  DE  =  (d+V/3);^C  =  d  _  r.     In 


,  (d   -  r)' 
2 


4±3\/2 


=  (-4  ±   3V2) 

VAC2-  CD2  = 
.'.  BC  =  .30d. 


BC  = 


-  .24 


=  AD  -   DG; 

;  ZX?  =  r\/3 


NOTE.  —  The  preceding  problems  are  found  in  "A  Source 
Book  of  Problems  for  Geometry,"  by  Mable  Sykes;  Allyn 
and  Bacon. 

322.  Various  Useful  Applications.  To  Draw  an  Arc 
Through  Three  Given  Points  When  the  Radius  Is  Large.  —  A 

B 
A 


pencil  point  at  B  will  describe  the  required  arc  if  the  arms 


CIRCLES 


233 


BA  and  BC  slide  along  pins  inserted  at  points  A  and  C. 
Why? 

To  Find  the  Diameter  of  a  Column. — A  carpenter  may  be 


required  to  cut  a  circular  column  along  a  diametral  plane. 
Explain  how  the  diameter  is  determined  by  laying  the 
square  on  the  column  as  shown. 

To  Make  a  Core  Box  Truly  Cylindrical.— 
Explain    how    a    pattern    maker    uses    his 

square,  and  why  the  vertex  of  the  square  ~- 

must  scrape  the  hollow  as  it  is  slid  around. 

To  Find  the  Radius  of  an  Arc. — Find  the  center  by  the 


method  of  constructing  a  circle  which  will  contain  three 
given  points. 

The  Length  of  a  Chord.  —  Show  how  r  is  ex- 
pressed in  terms  of  the  chord  c,  and  its  dis- 
tance d,  from  the  center.  When  d  =  $r,  show 
that  c  —  r\/3'  Solve  this  last  formula  also 
for  r  in  terms  of  c. 

The  Rise  of  an  Arc.  —  The  rise  of  the  arc  AB  from  the 
chord  c,  of  the  arc  is  a.     Show  that 
2r± 


c)(2r-c) 


_  d  ±  V(d  +  c)(d  -  c) 


Solve  also  for  r  or  d  in  terms  of  c  and  a.     r  = 


4a2 


8a 


234 


PLANE  GEOMETRY 


EXERCISES 

1.  Find  the  length  of  a  chord  6  inches  from  the  center  of  a  circle  of 
radius  12  inches. 

2.  Find  the  diameter  of  a  circle  if  a  chord  drawn  10  inches  from 
the  center  is  20  inches  long. 

3.  A  circular  arch  bridge  is  to  span  a  stream  52  feet  wide.     If  the 
rise  of  the  crown  of  the  arch  is  13  feet,  what  is  the  radius  of  the 
arch? 


4.  A  doorway  is  6  feet  wide.     The  rise  of 
the   circular    top   is    1    foot    above    the 
straight  sides.     Calculate  the  radius  of 
the  arc. 

5.  A     surveyor    wishes    to    find    the 
radius    of    a    railroad   curve.     He 
measures  a  chord  of  the  curve  = 
100   feet,    and    the   offset   at   the 
center  of  the  chord  =  2  feet.      What  is 
the  radius  of  the  curve? 

6.  A  chord  measured  on  a  piece  of  a  broken 
wheel  is  20  inches  and  the  rise  of  the  arc 
is  5  inches.     What  is  the  radius  of  the 
wheel? 

To  Lay  Out  a  Circular  Curve  by  Offsets  from  the  Tangent 

T 

d 


at  the  Beginning  Point  of  the  Curve. — Show  that  the  value  of 
the  perpendicular  offset  p,  at  a  distance  d,  from  the  point 
of  tangency  T,  is  p  =  r  —  \/r2  —  d2. 


CIRCLES  235 

EXERCISE 

7.  Calculate  the  offsets  from  the  tangent  of  a  curve  of  1000  feet 
radius,  at  distances  from  the  point  of  beginning  of  the  curve  of 
100,  200  and  300  feet. 

The   Strongest   Rectangular   Beam    That 


* 

so 


Can  Be  Cut  from  a  Circular  Log. — The 
beam  is  laid  out  on  the  end  of  the  log  by 
drawing  a  diameter  AB,  and  trisecting  at 
points  C  and  D,  and  erecting  perpendicu- 
lars CE  and  DF,  etc. 

EXERCISES 

8.  Calculate  the  sides  AE  and  BE  of  the  beam  in  terms  of  the 
diameter  d.     Show  that  ^^  =  ^  approximately. 

9.  What  are  the  dimensions  of  the  strongest  beam  that  can  be  cut 
from  a  log  2  feet  in  diameter? 

323.  PROBLEMS  FOR  FIELD  WORK. — (1)  To  Lay  Out  a 
Circular  Curve  by  Ordinates. — Use  the  data  of  319,  Exercise 
2,  or  calculate  the  ordinates  for  a  curve  of  any  other  radius. 
Stake  out  points  A,  C,  D,  -  —B;  erect  perpendiculars 
by  the  method  of  237;  measure  on  each  perpendicular  the 
length  of  the  calculated  ordinate;  drive  stakes  at  points 
C",  £>',  E',  etc. 

2.  To  Lay  Out  a  Circular  Curve  by  Offsets  from  the  Tan- 
gent.— Take  the  radius  of  the  curve  =  200  feet.  Select  a 
point  A  on  a  straight  line  where  the  curve  is  to  begin. 
Calculate  the  offsets  for  values  of  d  =  40,  80,  120  etc!  feet. 
See  322,  Exercise  7.  Stake  out  these  distances  along  the 
tangent  from  point  A,  and  erect  perpendiculars  to  the  tan- 
gent; measure  on  each  perpendicular  the  length  of  the 
calculated  offset  until  the  curve  has  turned  a  quarter  circle; 
extend  the  last  offset  which  is  the  tangent  at  the  other 
extremity  of  the  curve. 


CHAPTER  XIV 

REGULAR  POLYGONS.     CIRCLES  ASSOCIATED 
WITH  POLYGONS 

PRINCIPLES  DETERMINED  EXPERIMENTALLY 

324.  Previous  Mention. — Regular  polygons  have  been 
defined  and  briefly  considered  in  Chapter  VI;  inscribed  and 
circumscribed  polygons  in  Chapter  XII. 

325.  Experiment   I.    Equilateral  Inscribed  Polygons.— 
Draw  several  circles;  draw  equilateral  inscribed  polygons 

of  4,  5,  6,  7  sides  by  dividing  the  circumferences  by  trial 
into  equal  arcs. 

Observe  that  the  polygons  are  equiangular. 

EXERCISES 

1.  How  can  the  division  points  of  the  circumference  be  found  with- 
out trial  in  the  case  of  the  equilateral  quadrilateral? 

2.  How  can  the  division  points  be  found  without  trial  in  the  case 
of  the  hexagon? 

326.  Experiment  II.   Equiangular  Circumscribed  Poly- 
gons.— Draw    several    circles;    divide    the    circumferences 
into  4,  5,  6,  7  equal  arcs;  draw  polygons  whose  sides  are 
tangent  at  the  division  points.     Such  polygons  will  be 
equiangular  by  301. 

Observe  that  the  polygons  are  also  equilateral. 

327.  Experiment  III. — Properties  of  Regular  Polygons.— 
Calculate  each  interior  angle  of  regular  polygons  of  4,  5,  6, 

7  sides,  by  191,  Theorem  (5).     Draw  the  polygons  by  draw- 
ing alternately  a  side  and  an  angle  of  each  polygon. 

(a)  Bisect  all  the  interior  angles  of  the  polygons.     Ob- 
serve that  the  bisectors  are  concurrent. 

236 


REGULAR   POLYGONS  237 

(b)  Draw  the  perpendicular-bisectors  of  all  the  sides  of 
the  polygons. 

Observe  that  they  are  concurrent. 

(c)  Draw  the  inscribed  and  circumscribed  circles  of  all 
the  polygons. 

Observe  (1)  that  the  circles  of  each  polygon  are  concen- 
tric; (2)  that  the  inscribed  circle  is  tangent  to  each  side  at 
its  middle  point;  (3)  that  the  triangle  formed  by  a  half-side 
of  a  polygon,  a  radius  of  the  circumscribed  circle  and  a 
radius  of  the  inscribed  circle  is  a  right  triangle. 

328.  Definitions. — The  common  center  of  the  inscribed 
and  circumscribed  circles  of  a  regular  polygon  is  the  center 
of  the  polygon. 

The  radius  of  the  circumscribed  circle  of  a  regular  polygon 
is  the  radius  of  the  polygon. 

The  radius  of  the  inscribed  circle  of  a  regular  polygon  is 
the  apothem  of  the  polygon. 

The  right  triangle  formed  by  a  half-side,  radius  and  apo- 
them is  an  element  of  the  polygon. 

EXERCISES 

1.  Calculate  the  central  angle  of  an  element  of  a  regular  polygon  of 
4,  5,  6,  7,  sides. 

2.  Write  a  general  formula  for  the  central  angle  of  an  element  of  a 
regular  polygon  of  n  sides. 

3.  Write  a  formula  for  a  half -side  s,  expressed  in  terms  of  the  per- 
imeter p. 

4.  Express  a  general  relation  between  a  half -side  s,  a  radius  r,  and 
an  apothem  a. 

5.  Write  a  general  formula  for  one  of  the  equal  angles  of  a  triangle 
formed  by  two  adjacent  sides  of  the  polygon  and  a  diagonal  join- 
ing the  extremities  of  the  two  sides. 

Help. — (1)  Write  the  value  of  one  interior  angle  of  the  polygon; 

180° 
(2)  subtract  from  180°;  (3)  divide  by  2.     Value  is  • 

329.  PROBLEMS  RELATING  TO  REGULAR  POLYGONS. — In 

these  problems  only  an  element  of  the  polygon  is  required, 
except  in  those  which  involve  diagonals. 


238  PLANE    GEOMETRY 

1.  Find  by  measurement  the  radius  and  apothem  of  a 
regular  pentagon  of  which  the  side  is  1.4  inches. 

2.  Find  by  measurement  the  side  and  radius  of  a  regular 
hexagon  of  which  the  apothem  is  1.2  inches. 

3.  Derive   a   formula   for  the  value  of  the  side  and 
radius  of  an  equilateral  hexagon,  expressed  in  terms  of 
the  apothem. 

Help.— Refer  to  260,  Theorem  (18). 

4.  Check  the  measured  values  of  Exercise  2  by  calcu- 
lating the  same  values  by  the  formula  of  Exercise  3. 

5.  Find  by  measurement  the  apothem  and  radius  of  a 
regular  pentagon  of  which  the  perimeter  is  30  feet. 

6.  Find  by  measurement  the  apothem  and  side  of  a 
regular  decagon  of  which  the  radius  is  2  inches. 

7.  Find  the  number  of  sides  of  a  regular  polygon  of  which 
the  side  is  2  inches  and  radius  2.61  inches. 

Helps. — Construct  the  element;  measure  the  central 
angle;  divide  it  into  360°. 

8.  Find  the  number  of  sides  of  a  regular  polygon  of 
which  the  side  is  20  feet  and  the  apothem  37.32  feet. 

9.  The  least  diagonal  of  a  regular  decagon  is  2  inches. 
Find   by  measurement  the  side,  apothem  and  radius. 

Helps. — (1)  Calculate  the  angles  of  the  triangle  formed 
by  diagonal  and  two  adjacent  sides,  by  328,  Exercise  5;  (2) 
Construct  this  triangle.  (3)  Construct  the  element  of  the 
polygon  on  the  side  as  thus  determined. 

10.  The  least  diagonal  of  a  regular  polygon  is  2.83  inches 
and  a  side  is  2  inches.     Find  the  number  of  sides  of  the 
polygon. 

11.  Calculate  the  radius,  apothem  and  perimeter  of  a 
square  of  which  the  side  is  20  inches. 

12.  Calculate  the  radius,  apothem  and  perimeter  of  an 
equilateral  triangle  of  which  a  side  is  20  inches. 

13.  Calculate  the  apothem  and  radius  of  a  regular  octagon 
of  which  the  side  is  6  inches. 

Helps.— (1)  62  =  r2  +  r2  -  2r(.707r)  (252  and  254);  (2) 
.586r2  =  36;  r  =  \/61.43  =  7.84;  (3)  a2  =  (r  +  s}  (r  -  s). 


REGULAR   POLYGONS 


239 


330.  EXPERIMENT  IV. — THE  RATIO  OF  THE  CIRCUMFER- 
ENCE TO  THE  DIAMETER  OF  A  CIRCLE. 

(a)  Draw  a  number  of  circles  of  different,  size;  say  1.0", 
1.5",  2.0",  2.5",  3.0"  in  diameter.  Measure  the  circumfer- 
ences as  carefully  as  possible  with  a  flexible  scale.  A  narrow 
uniform  strip  of  paper  placed  on  edge  may  be  used.  Enter 
the  values  in  column  c'  of  the  accompanying  table. 


d 

c' 

c" 

c'" 

Average 
value 
of  c 

It" 

Average 
value 
of* 

1.0 

1.5 

2.0 

2.5 

3.0 

(6)  Rankine's  Approximation  for  Finding  the  Length  of  a 
Circular  Arc. — AB  is  the  given  arc  with  center  C.     Draw 


A      F  GO 

radius  CA  and  tangent  AD.  Find  AE  =  J  arc  AB,  and 
draw  arc  EF  with  center  A.  Draw  arc  BG  with  center  F. 
Then  sect  AG  =  arc  AB}  approximately.  The  error  in  this 
method  almost  vanishes  for  small  arcs  and  reaches  ¥^  of 
the  length  of  the  arc  for  an  arc  of  60°.  It  should  not  be  used 
for  arcs  exceeding  90°. 

Use  the  circles  of  part  (a)  and  find  the  lengths  of  arcs  of 
60°  for  each  circle.  Multiply  by  6  and  enter  the  values 
in  column  c"  of  the  table. 

(c)  Ceradini's  Approximation. — Draw  diameter  AB;  tan- 
gent AD;  ^ACE  =  30°;  make  EF  =  3r;  draw  BF.  Then 


240 


PLANE    GEOMETRY 


BF  =  arc  AB  =  J  circumference.     Find  the  values  of  the 
circumferences  of  the  circles  used  in  Parts  (a)  and  (6)  by  this 
method.     Enter  in  column  c"r  of  the  table. 
B 


E          A 
Complete  the  table  as  indicated. 

EXERCISES 

1.  Calculate  the  circumference  of  a  circle  of  which  the  diameter  is 
20  feet. 

2.  Calculate  the  diameter  of  a  circle  of  which  the  circumference  is 
50  feet. 

3.  What  is  the  diameter  of  a  cask  if  the  circumference  measures 
10  feet?     Will  the  cask  pass  through  a  door  3  feet  wide? 

4.  WTrite  a  formula  for  the  length  of  an  arc  of  F  degrees,  in  a  circle 
whose  diameter  is  d. 

331.  THE  LENGTH  OF  AN  ARC  DETERMINED  FROM  ITS 
CHORD. — A  formula  that  gives  the  length  of  an  arc  not  ex- 
ceeding 60°,  with  close  approximation  is  I  =  — « — /  where 

C  is  the  chord  of  the  arc  and  c  is  the  chord  of  J  the  arc. 

To  use  the  formula,  construct  the  arc  to  scale  and  measure 
the  chords. 

EXERCISES 

1.  Find  the  length  of  an  arc  of  45°  in  a  circle  whose  radius  is  20 
inches. 

2.  Calculate  the  length  of  the  arc  of  Exercise  1  by  using  the  formula 
of  330,  Exercise  4. 

3.  The  chords  of  Exercise  1  may  be  calculated  as  follows:  (1)  C*  = 

(C\  2 
2 )  >  where    a 


REGULAR   POLYGONS  241 

is   the   apothem  of   a   regular   octagon    whose  side   is    C;  (3) 

40  c 

—  =  2Q  _ —     Substitute    these    values    in    the   formula  I  = 

8c-C 
— n — »  and  solve  for  I. 


332.  Exact  Constructions  of  Regular  Inscribed  Polygons. 
The  problem  of  finding  rule  and  compass  constructions 
for  regular  inscribed  polygons,  has  been  a  favorite  one 
with  mathematicians.  Methods  have  been  given  in 
preceding  chapters  for  the  rule  and  compass  construction 
of  inscribed  equilateral  triangles,  squares  and  regular 
hexagons.  By  successively  bisecting  the  arcs  subtended 
by  the  sides  of  these  polygons,  regular  inscribed  polygons 
of  8,  12,  16,  24,  32,  etc.,  sides  may  be  constructed. 

A  Rule  and  Compass  Construction  of  a  Regular  Inscribed 
Decagon. — Draw  a  radius  OA;  divide 
it  into  mean  and  extreme  ratio  at 
point  B  (316);  OB  is  the  required  side 
of  the  decagon.  A  regular  inscribed 
pentagon,  and  regular  polygons  of 
20,  40,  etc.,  sides  can  be  obtained 
from  the  decagon. 

For    all    practical    purposes,    the 
trial  and  error  method  used  in  325 
is  the  best,  except  for  polygons  of  3,  6,  12,  etc.,  and  4,  8, 
16,  etc.,  sides. 

In  1796,  Gauss,  the  greatest  mathematician  of  the 
Eighteenth  century,  then  nineteen  years  of  age,  discovered 
a  method  of  constructing  all  regular  inscribed  polygons 
the  number  of  whose  sides  is  given  by  the  formula 

n  =  2*(22"  +  1) 
where  p  and  q  are  integers. 

EXERCISE 

Find  the  value  of  n  in  Gauss'  formula  for  q  =  0,  p  =  0;  q  =  0, 
p  =  1;  q  =  1,  j)  =  0:  a  =  1,  p  =  1;  etc.,  to  q  =  3,  p  =  3. 
16 


242  PLANE   GEOMETRY 

APPLICATIONS 

333.  (1)  An  hexagonal  nut,  one  inch  on  each  side,  is  to  be 
cut  from  a  round  bar  of  iron.  What  diameter  of  stock  must 
be  taken? 

2.  A  square  nut  is  to  be  made  for  a  bolt  1J  inches  in 
diameter.     The  least  thickness  of  metal  around  the  hole 
equals  the  radius  of  the  bolt.     What  diameter  of  round 
stock  must  be  taken? 

3.  A  gardener  plans  a  summer-house  to  be  laid  out  as  a 
regular  octagon.     The  radius  is   10  feet.  /, 

He  drives  a  stake  C,  at  the  center  and  a  £> 
second  stake  A  at  the  center  of  the  arc 
subtended  by  the  side  parallel  to  the  path. 
Measure  from  a  drawing  of  an  element 
made  to  scale,  the  chords  AB  and  (side) 
BD.  The  stakes  B,  D,  E,  etc.,  can  then 
be  placed. 

4.  In  the  description  of  Solomon's  Temple   (1   Kings, 
VII,   23)    the   dimensions   of   the   " Molten   sea,"   which 
appears  to  have  been  a  great  circular  brazen  vessel,  are 
given  as  "ten  cubits  from  the  one  brim  to  the  other" 
(diameter),  and  "a  line  of  thirty  cubits  did  compass  it 
round   about"    (circumference).     What   was  the  writer's 
estimate  of  the  value  of  TT? 


CHAPTER  XV 
REGULAR  POLYGONS  AND  CIRCLES 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

334.  Constants,  Variables  and  Limits. — A  constant  is  a 
quantity  which  possesses  a  fixed  value. 

A  variable  is  a  quantity  whose  value  changes  continually. 

The  limit  of  a  variable  is  the  constant  which  the  variable 
continually  approaches,  but  may  never  quite  equal. 

EXERCISES 

1.  One  half  of  a  pile  of  sand  is  removed  in  a  day,  one  half  of  the 
remainder  is  removed  the  second  day,  and  so  on.     The  whole 
pile  of  sand  is  a  constant  value;  the  total  amount  removed  up 
to  any  given  tune  is  a  variable.     What  is  the  limit  of  the  variable? 

2.  The  sum  of  the  infinite  geometrical  series,  $  +  i  +  i  +,  etc., 
for  n  terms,  is  a  variable.     What  is  the  limit  of  this  variable  as 
n  is  indefinitely  increased? 

3.  What  is  the  limit  of  the  repeating  decimal,  .333  .    .   .    .   ? 

Analysis  by  the  Principles  of  Limits. — Some  analyses 
involving  metrical  relationships  (that  is,  where  numerical 
values,  or  measures,  are  involved)  are  made  by  obtaining  a 
relation  under  certain  given  conditions,  and  then  considering 
the  figure  to  undergo  successive  modifications  until  it 
merges  into  another  figure.  If  the  property  which  was 
shown  to  exist  in  the  primary  figure  holds  true  during  the 
successive  modifications  of  that  figure,  it  holds  true  in  the 
final  figure.  This  final  figure  is  the  limit  of  the  changing 
or  variable  figure;  the  values  of  sects,  angles,  arcs,  areas, 
ratios,  in  the  final  figure  are  the  limits  of  the  variable  sects, 
angles,  etc.,  of  the  variable  figure. 

243 


244  PLANE    GEOMETRY 

EXERCISE 

4.  If  a  regular  polygon  possesses  some  property  which  remains  un- 
changed in  principle  as  the  number  of  sides  of  the  polygon  is 
indefinitely  increased,  what  limiting  figure  must  also  possess 
this  same  property? 

Abbreviation. — Approaches  as  a  limit,  ==. 
THE  FIRST  STEP  IN  CLASSIFICATION 

335.  Postulates    of   Regular   Polygons   and   Circles.— 

(1)  An  equilateral  inscribed  polygon  is  regular. 

2.  A  circumscribed  polygon  whose  points  of  tangency 
divide  the  circumference  of  the  circle  into  equal  arcs,  is 
regular. 

3.  The  perimeter  of  a  regular  inscribed  polygon  (a)  is 
less  than  the  circumference  of  the  circle;  (6)  increases  as  the 
number  of  sides  is  indefinitely  increased;  (c)  approaches  the 
circumference  of  the  circle  as  a  limit. 

4.  The   perimeter  of  a  regular   circumscribed  polygon 
(a)  is  greater  than  the  circumference  of  the  circle;  (6)  de- 
creases as  the  number  of  sides  is  indefinitely  increased;  (c) 
approaches  the  circumference  of  the  circle  as  a  limit. 

5.  Two  regular  polygons  of  the  same  number  of  sides  are 
similar  (see  234). 

6.  Corresponding  sects  of  two  regular  polygons  of  the 
same  number  of  sides  are  proportional. 

Limit  Axioms. — (1)  If  two  variables  are  proportional  to 
two  constants,  the  limits  of  the  variables  are  proportional 
to  the  constants. 

2.  If  two  variables  approaching  limits  are  always  equal, 
their  limits  are  equal. 

3.  If  a  variable  is  approaching  a  constant  as  a  limit,  the 
variable  multiplied  by  or  divided  by  any  number  approaches 
the  constant  multiplied  by  or  divided  by  the  same  number, 
as  a  limit. 

EXERCISES 

1.  How  much  less  is  the  perimeter  of  a  regular  hexagon  inscribed 
in  a  circle  whose  radius  is  10  inches,  than  the  circumference  of 
the  circle?  Use  the  value  of  TT  =  3|. 


REGULAR   POLYGONS   AND    CIRCLES 


245 


2.  How  much  greater  is  the  perimeter  of  a  regular  quadrilateral 
circumscribed  about  a  circle  whose  radius  is  10  inches,  than  the 
circumference  of  the  circle? 

3.  If  the  diagonal  of  a  square  whose  side  is  10,  is  14.142,  what  is 
the  diagonal  of  a  square  whose  side  is  20? 

4.  The  apothem  of  a  regular  hexagon  whose  side  is  10,  is  8.66. 
What  is  the  apothem  of  a  regular  hexagon  whose  side  is  30? 
What  is  the  side  of  a  regular  hexagon  whose  apothem  is  34.64? 

THE  SECOND  STEP  IN  CLASSIFICATION 

336.  Theorem  I. — -The  perimeters  of  two  regular  polygons 
of  the  same  number  of  sides  are  proportional  to  the  radii  of  the 
polygons. 


Hypothesis. — Regular  polygons  0  and  0',  of  the  same 
number  of  sides. 

Conclusion. —  —.  =  ->• 
p'      r' 

Analysis. 


STATEMENT 
AB        BC        CD 
A'B'~  B'C'  ~  C'D' 


REASON 
335;  Postulate  5 


2. 


AB    +  BC  +  CD  +etc. 


A'B'  +  B'C'  +  C'D'  +  etc. 
3-  Or?/  =  -i7D> 


4. 


AB 
A'B1 


*      .     P  _r 
5*   "    p'-r' 


64;    Axioms    11 
and  13 

335;  Postulate  6 
64;  Axiom  1 


246 


PLANE    GEOMETRY 


337.  Theorem  II. — The  ratio  of  the  circumference  to  the 
radius  is  constant  for  all  circles.  Or:  Circumferences  of 
circles  are  proportional  to  their  radii. 


Hypothesis. — 0  and  0'  are  circles  with   circumferences 
c  and  c',  and  radii  r  and  rr. 

c        c' 

Conclusion. =  — ,• 


Analysis. 


STATEMENT 

1.  Inscribe  regular  polygons  of  the 
same  number  of  sides  in  the  two 
circles,  the  perimeters  being  p 
and  p1 '. 

2.  The  polygons  are  similar. 

3.  Let  the  number  of  sides  of  the 
polygons    be    indefinitely    in- 
creased by  doubling  the  num- 
ber of  sides;  then  the  successive 
polygons  are  regular,  and  simi- 
lar at  each  step. 

p       r 

4-  p>  =  ?• 

5.  But  p  =  c,  and  p'  =  c'. 


REASON 


— .  =  -.  or  -  = 
c'     r'      r 


335;  Postulates  1 
and  5 


335;  Postulates  1 
and  5 

336;  Theorem  I 
335 ;  Postulate  3 
335 ;  Limit  Axiom  1 


KEGULAR   POLYGONS   AND    CIRCLES 


247 


Corollaries.  —  (1)    The  ratio  of  the  circumference  to   the 

c      cr 

diameter  is  constant  for  all  circles.     That  is;  -3  =  -r,  =  etc.  =  TT. 

2.  In  any  circle;  (a)  c  =  nd;  (b)  c  =  2irr. 

3.  The  length  of  an  arc  of  n  degrees  of  a  circle  whose  radius 

is  r,  is  a  =         -  2irr. 


THE  THIRD  STEP  IN  CLASSIFICATION 


338.  Theorem  III.—  The  side  of  a 
regular  inscribed  polygon  can  be  ex- 
pressed in  terms  of  the  side  and  radius 
of  the  regular  inscribed  polygon  of  half 
the  number  of  sides. 

Hypothesis.  —  AC  is  the  side  x,  of  a 
regular  inscribed  polygon  of  n  sides; 
A  B  is  the  side  s  of  a  regular  inscribed 

polygon  of     sides;  R  is  the  radius. 


Conclusion.  —  x  may  be  expressed  in  terms  of  s  and  R. 
Analysis. 

STATEMENT 

1.  Draw  CD,  OA,  AD 

2.  CD  J_  AB 

3.  AE  =  |- 

4.  Z.CAD  is  a  right   Z. 

5.  x2  =  CD-CE 


6.  x2  =  2R(R  -  OE) 


R  - 


7.  x2  =  R(2R  - 


8.  x    =VR(2R  - 
Corollary.—  If  R  =  1,  x 


-  s2). 


248 


PLANE    GEOMETRY 


EXERCISE 

The  side  of  a  regular  hexagon  inscribed  in  a  circle  of  unit  radius  is  1. 
Calculate  the  side  of  a  regular  dodecagon  (12  sides)  inscribed  in  the 
same  circle. 

339.  Theorem  IV.—  The  side  of  a 
regular  circumscribed  polygon  can  be 
expressed  in  terms  of  the  side  and 
apothem  of  the  regular  circumscribed 
polygon  of  half  the  number  of  sides. 

Hypothesis— FC,  CD,  are  the  sides 
x,  of  a  regular  circumscribed  polygon 
of  n  sides;  AB  is  the  side  s  of  a 

regular  circumscribed  polygon  of  ~ 
sides;  r  is  the  apothem. 

Conclusion. — x  may  be  expressed  in  terms  of  s  and  r. 

Analysis. 

STATEMENT 
1.  Draw  OE,  OC,  OB 


2.  OC  bisects 
EC  =  OE 
CB      OB 
EC 


Z.EOE. 


3. 


OE 


L'  EB  -  EC 

x 

5        *     - 

\/OE2  + 
r 

EB* 

2  ~  2     V 
6.  *  - 

"+©' 

2rs 

2r  +  V  4r2  +  s2 
Corollary . — If  r  =  1,  x  = 


2s 


2  +  -v/4  +  s2 
EXERCISE 

The  side  of  a  regular  quadrilateral  (square)  circumscribed  about  a 
circle  of  unit  radius  is  two.  Calculate  the  side  of  a  regular  octagon 
circumscribed  about  the  same  circle. 


REGULAR   POLYGONS   AND    CIRCLES 


249 


340.  Theorem  V. — The  value  of  IT  can  be  calculated  with 
as  close  an  approximation  as  desired. 

,0 


Hypothesis. — Circle  0,  with  radius  r  =  1;  circumference 
c;  diameter  d  =  2r  =  2. 

C  C 

Conclusion. — The  value  of  r  =  -3  or  ~~  can  be  calculated, 

etc. 
A  nalysis. 

STATEMENT  REASON 

1 .  Draw  a  regular  inscribed  hexagon 
with  side  AB  =  s,  and  a  circum- 
scribed square  with  side  CD  —  s'. 

2.  Side  AB  =  s  =  r  =  1;  side  CD  = 
s'  =  2r  =  2 

3.  The  perimeter  of  the  hexagon  = 
6;  the  perimeter  of  the  square  =  8 

4.  .*.  c  >6  and  <8;   .'.  TT>!  or  3  335;  Postulates 3 and 4 
and  <  f  or  4 

5.  The  perimeter  of  a  regular  in- 
sribed   polygon   of    12   sides  = 

12[\/2^  A/4^T|  =6.211657       338;  Theorem  III 
The  perimeter  of  a  regular  cir- 


250 


PLANE    GEOMETRY 


cumscribed    polygon    of  8  sides 

2  2 

627418        339;  Theorem  IV 


1 
']  =  6-' 


f  A/4  +  22 
6.  .'.  c  >  6.211657  and  <  6.627418 

6<2101657or  3. 105829  and  < 


or  3.313709 


6.627418 


335;  Postulates  3  and  4 


7.  If  the  doubling  of  the  sides  of 
inscribed  and  circumscribed  poly- 
gons is  continued  indefinitely,  the 
following  tables  are  obtained: 


REGULAR  INSCRIBED  POLYGONS 


Ratio  of 

Ratio  of 

No.  of 
sides 

Perimeter 

perimeter 
to  diameter 

No.  of 
sides 

Perimeter 

perimeter 
to  diameter 

of  circle 

of  circle 

6 

6 

3 

4 

8 

4 

12 

6.211657 

3.105829 

8 

6.627418 

3.313709 

24 

6.265257 

3.132629 

16 

6.365196 

3.182598 

48 

6.278700 

3.139350 

32 

6.303450 

3.151725 

96 

6.282064 

3.141032 

64 

6.288226 

3.144118 

192 

6.282905 

3.141453 

128 

6.284448 

3.142224 

384 

6.283115 

3.141558 

256 

6.283500 

3.141750 

512 

6.283264 

3.141632 

REGULAR  CIRCUMSCRIBED 
POLYGONS 


8.  The  perimeters  of  both  inscribed 
and  circumscribed  regular  poly- 
gons approach  the  circumference 
c  as  a  limit: 


9. 


and  *L 


C 

-j  or 
a 


'  d"    4  d 

>  3.141558  and  <  3.141632 
TT  =  3.1416,  approximately. 


335;  Limit  Axiom  3 


REGULAR   POLYGONS    AND    CIRCLES*  251 

341. — The  History  of  TT. — The  determination  of  TT  is  one 
of  the  famous  geometric  problems.  It  is  attempted  in  the 
MS.  of  Ahmes,  now  preserved  in  the  British  Museum. 
Other  later  attempts  took  the  form  of  a  determination  of 
the  area  of  a  circle,  or  of  the  side  of  a  square  of  equal  area. 
Such  a  value  involves  that  of  TT,  and  thus  the  problem  has 
been  known  as  "squaring  the  circle." 

Archimedes  (250  B.C.),  and  Ptolemy  (150  A. D.),  obtained 
values  of  TT  almost  as  accurate  as  that  calculated  in  340. 
Metrus  and  Romanus,  of  Holland,  both  belonging  to  the 
16th  century,  calculated  the  value  of  TT  by  the  method  of 
340,  the  latter  using  polygons  up  to  1,073,741,324  sides, 
and  obtaining  a  value  correct  to  sixteen  decimals.  Lam- 
bert (1750)  proved  TT  incommensurable,  that  is,  having  no 
exact  value;  and  Lindemann  (1882)  proved  it  transcendental, 
that  is,  not  expressible  as  a  radical  nor  as  a  root  of  an  alge- 
braic equation  with  integral  coefficients.  There  are  other 
methods  of  calculating  TT,  which  belong  to  the  realm  of 
higher  mathematics.  Mr.  Shanks,  in  1873,  carried  the 
calculation  to  707  decimals.  Ten  decimals  are  necessary 
for  the  most  exact  astronomical  calculations,  and  four  to 
seven  decimals  are  sufficient  for  those  of  mechanics  and 
engineering. 

The  establishment  of  the  theorem  of  340  may  justly  be 
regarded  as  one  of  the  greatest  triumphs  of  the  logical 
system. 

The  number  r  enters  into  a  great  many  important  mathe- 
matical expressions,  or  formulas,  which  have  nothing  what- 
ever to  do  with  a  circle.  The  fact  that  TT  equals  the  ratio 

^  is  really  not  its  definition,  but  should  be  regarded  merely 

as    an    incidental — and    the   most    elementary — property 
(Ball's  "  Mathematical  Recreations  and  Problems"). 

ADDITIONAL  THEOREMS 

342.  (1)  The  apothem  and  radius  of  a  regular  triangle 
can  be  expressed  in  terms  of  the  side. 


252  PLANE    GEOMETRY 

Helps.—  (a)    r  =  =  .2887s; 


(6)  R  =  =  .5774s. 

o 

(2)  The  side  of  a  regular  triangle  can  be  expressed  in  terms 
of  (a)  the  apothem,  and  (b)  the  radius. 

Helps.—  (a)  s  =  2r\/3  =  3.4642r: 
(6)  s  =  #V3  =  1.7321#. 

(3)  The  central  angle  subtended  by  a  side  of  a  regular 
polygon  is  double  the  angle  between  a  side  and  a  least  diagonal. 

(4)  The  perimeter  of  the  circumscribed  equilateral  triangle 
of  a  circle  has  a  fixed  ratio  to  the  perimeter  of  the  inscribed 
equilateral  triangle. 

(5)  Find  the  ratio  of  the  sides  and  perimeters  of  the  inscribed 
and  circumscribed  squares  of  a  circle. 

(6)  Find  the  ratio  of  the  sides  and  perimeters  of  the  in- 
scribed and  circumscribed  regular  hexagons  of  a  circle. 

(7)  The  side  of  a  regular  decagon  whose  radius  is  r,  is, 


Helps—  By  332,  —  —  =  -•    See  also  316. 
s          r 

343.  THE  FOURTH  STEP  IN  CLASSIFICATION. 


EXERCISES 

1.  Draw  a  regular  inscribed  polygon  of  5  sides,  in  a  circle  whose 
radius  is  about  3  inches.     Bisect  one  of  the  arcs  subtended  by 
a  side  of  the  polygon  and  join  the  bisection  point  with  an  end 
of  the  nearest  side,thus  obtaining  a  side  of  the  regular  inscribed 
polygon  of  10  sides.     Measure  a  side  of  each  polygon.     Calcu- 
late the  side  of  the  decagon  from  the  side  of  the  pentagon  by 
using  the  formula  of  338.     Does  the  calculated  result  agree 
with  the  measured  value? 

2.  Use  the  figure  of  Exercise  1  and  draw  a  regular  circumscribed 
pentagon  and  one  side  of  the  regular  circumscribed  decagon. 
Measure  a  side  of  each  polygon.     Calculate  the  side  of  the 
decagon  from  that  of  the  pentagon  by  using  the  formula  of  339. 
Does  the  calculated  result  agree  with  the  measured  value? 


REGULAR   POLYGONS   AND   CIRCLES  253 

3.  Construct  accurate  figures  for  all  the  Theorems  of  342  and  test 
the  correctness  of  the  formulas  and  other  principles  of  the 
section  by  careful  measurements  of  the  figures.  The  figure  of 
Exercise  1  may  be  used  for  the  test  of  Theorems  (3)  and  (7). 

344.  REVIEW  EXERCISES 

.1.  If  a  string  6  feet  long  is  laid  so  as  to  form  an  equilateral  triangle, 
what  are  the  radii  of  the  inscribed  and  circumscribed  circles? 

2.  Can  a  cask  which  measures  9  feet  in  circumference  be  taken 
through  a  3-foot  door? 

3.  Calculate  the  length  of  an  arc  of  75°  in  a  circle  whose  radius  is 
1200  feet. 

Helps. — (a)  Use  the  formula  of  337,  Corollary  3;  (b)  use  the  for- 
mula of  331,  the  chords  C  and  c  being  measured  from  a  figure 
drawn  to  scale. 

4.  The  length  of  an  arc  of  60°  is  12.35  inches.     Calculate  the  diame- 
ter of  the  circle. 

5.  The  circumference  of  a  circle  is  10  feet.     What  is  the  circum- 
ference of  a  circle  of  double  the  radius? 

6.  The  chord  C  of  an  arc  of  60°  in  a  circle  whose  radius  is  10  feet, 
is  10.     Calculate  the  length  of  an  arc  of  60°  by  the  formula  of 
337,  Corollary  3.     Also  (a)  calculate  the  chord  c  of  one-half  the 
arc  by  the  formula  of  331 ;  and  (6)  calculate  this  chord  by  the 
formula  of  338 ;  and  (c)  find  some  other  method  of  calculating 
this  chord. 

7.  Find  the  diameter  of  a  baseball  or  tennis  ball  by  measuring  the 
circumference. 

8.  Calculate  the  side  of  an  equilateral  triangle  of  which  the 
apothem  is  30. 

9.  Find  the  length  of  an  arc  of  60°  in  a  circle  whose  radius  is  30 
inches  (337,  Corollary  3). 

10.  Find  the  length  of  an  arc  of  45°  in  a  circle  whose  radius  is  20 
inches. 

11.  Find  the  length  of  an  arc  of  115°  35'  in  a  circle  whose  diameter 
is  50  inches. 

lt,35 

n  60  _    6,935 

(eLp'     360  ~      360         21,600 

12.  The  chord  C  of  an  arc  of  60°  in  a  circle  whose  radius  is  30  inches 
is  30  inches.     Calculate  the  chord  of  \  the  arc  from  this  value 
and  the  result  obtained  for  the  length  of  the  arc  in  Exercise  9; 
(a)  using  the  formula  of  331 ;  (b)  using  the  formula  of  338. 

13.  If  the  diagonal  of  a  square  whose  side  is  5,  is  7.07,  what  is  the 
diagonal  of  a  square  whose  side  is  8?     What  is  the  side  of  a 
square  whose  diagonal  is  8  (335,  Postulate  5)  ? 


254 


PLANE    GEOMETRY 


APPLICATIONS 

345.  The  Design  of  a  Running  Track. — The  best  and 
usual  shape  of  a  running  track  is  that  of  a  rectangle  with 
semi-circular  ends  (Fig.  1).  The  form  of  Fig.  2  may  be 
used  when  the  available  ground  space  does  not  permit  the 
use  of  the  better  form.  The  standard  line  on  which  the 
length  of  the  track  is  calculated  is  indicated  by  the  dotted 
line,  and  is  1|  feet  from  the  inner  edge  of  the  track. 


FIG.  1. 


FIG.  2. 


EXERCISES 


1.  Calculate  the  total  length  of  a  track  if  AB  =  200  feet  and  BC  = 
200  feet  (Fig.  1). 

Help. — The  two  ends  form  a  complete  circle  203  feet  in  diameter. 

2.  Calculate  the  total  length  of  a  track  of  the  form  of  Fig.  2,  if 
AB  =  200  feet,  BC  =  200  feet,  and  the  radius  of  the  corner 
turns,  measured  to  the  standard  line,  =  50  feet. 

Helps.— EF  =  203  -  2  X  50  =  103  feet. 

3.  What  must  be  the  length  of  AB  for  a  quarter-mile  track  if  GK 
(Fig.  1)  =  200  feet? 

4.  What  must  be  the  value  of  BC  for  an  eight-laps-to-the-mile  track, 
in  which  AB  =  100  feet  (Fig.   1)?     Find  the  over-all  length, 
/«/,  to  the  standard  line. 

5.  The  over-all  length,  /«/,  of  a  quarter-mile  running  track  is  400 

feet.     Find  AB  and  GK  (Fig.  1). 
Helps.— Let  x  =  GK;  then  AB  =  400  -  x;  .'.  length  of  track  is 

2(400  -  a)  +  3.1416*  =  ^~- 

6.  How  many  feet  more  will  a  runner  travel  in  one  lap  of  the  quar- 
ter-mile track  of  Exercise  3,  if  he  runs  3  feet  from  the  inner  edge 
of  the  track,  than  he  would  travel  on  the  standard  line? 

346.  POWER  TRANSMISSION  AND  SHOP  PRACTICE. — The 
simplest  forms  of  power  transmission  is  by  gear  wheels  and 
belts.  The  velocity  ratio  of  the  driving  shaft  A,  and  the 


KEGULAR   POLYGONS   AND    CIRCLES 


255 


driven  shaft  B,  is  the  ratio  of  the  cir- 
cumferences of  the  pitch  circles  of  the 
two  gear  wheels;  since  the  point  of 
contact  of  the  teeth  of  the  gear  wheels 
is  at  the  point  of  tangency  of  these 
circles;  Fig.  1.  The  velocity  ratio  of 
the  two  shafts  C  and  D  is  the  ratio  of 
the  circumference  of  the  driving  pulley 
C  and^the  driven  pulley  D,  around 
which  the  belt  passes;  Figs.  2  and  3. 


FIG.  1. 


EXERCISES 


1.  The  diameter  of  gear  wheel  A  (Fig.  1)  is  12  inches,  and  that 
of  gear  wheel  B  is  8  inches.     The  number  of  revolutions  per 
minute  of  shaft  A  is  60.     How  many  revolutions  per  minute 
does  shaft  B  make? 

2.  What  is  the  velocity  ratio  of  the  first  and  third  shafts  of  a  train 
of  three  gears  of  10,  12  and  8  inches  diameter,  respectively? 


FIG.  2. 


FIG.  3. 


3.  The  driving  gear  of  a  bicycle,  carried  on  the  pedal  shaft,  is 
8  inches  diameter;  the  rear  gear  on  the  wheel  axle  is  3  inches 
diameter;  the  tire  of  the  rear  wheel  is  28  inches  diameter. 
How  many  revolutions  per  minute  must  the  pedals  make  in 
order  to  drive  the  bicycle  at  the  rate  of  15  miles  per  hour? 
Help. — Number  of  revolutions  per  minute  of  pedal  shaft  = 

5280  X  15  X  12 1 3 

60  X  28  X  3.1416       8* 

4.  If  the  diameters  of  the  pulleys  on  both  shafts  in  Fig.  2  are  24 
inches,  and  the  shafts  are  10  feet  apart,  center  to  center,  calcu- 
late the  length  of  belt  required. 

5.  In  Fig.  2,  the  diameters  of  the  pulleys  are  36  inches  and  18  inches, 
and  the  distance  of  the  shafts  is  8  feet,  center  to  center.     Draw 
the  pulleys  and  the  distance  between  centers  to  scale,  measure 
the  lengths  SM  =  PQ,  and  the  angles  MDP  and  QCS,  and 


256 


PLANE    GEOMETRY 


calculate  the  arcs  MNP  and  QRS  by  the  method  of  337. 
Find  the  total  length  of  the  belt. 

6.  Use  the  dimensions  of  Exercise  5  to  apply  to  the  crossed  belt 
of  Fig.  3.     Calculate  the  length  of  the  belt. 

The  cutting  speed  of  a  tool  or  the  velocity  of  some  moving 
part  of  a  machine  must  be  known  by  those  who  design 
machinery. 

EXERCISES 

7.  If  the  allowable  cutting  velocity  of  a  tool  when  a  piece  of  iron  is 
turned  in  a  lathe,  is  40  feet  per  minute  at  the  surface  of  the 
revolving  piece,  at  how  many  revolutions  per  minute  should  a 
lathe  be  driven  in  order  to  produce  this  surface  velocity  for  a 
piece  of  iron  which  is   5  inches  in  diameter?     d  inches  in 
diameter? 

•8.  At  what  velocity  does  the  rim  of  a  10-foot  diameter  fly-wheel 
travel,  if  it  makes  80  revolutions  per  minute? 

9.  If  pulley  C  in  Figs.  2  and  3  is  24  inches  diameter  and  makes 
50  revolutions  per  minute,  what  is  the  linear  velocity  of  the  belt 
in  feet  per  minute? 

10.  If  a  belt  is  moving  at  a  linear  velocity  of  500  feet  per  minute, 
and  is  driving  a  shaft  by  a  3-foot  pulley,  how  many  revolutions 
per  minute  is  the  driven  shaft  making? 

Draughting  and  other  applications. 

EXERCISES 

11.  Show  that  in  drawing  a  side  view  of  a  hexagonal 
nut,  that  AB  =  CD  =  Id,  and  BC  =  \d.     In 
practice,  d  is  made  2di. 

12.  Draw  a  bolt  whose  diameter  di  is  1£  inches,  and 
a  hexagonal  nut. 

13.  ABCDis  a  square.     With  centers  at  A,  B,  C,  D, 
arcs  are  drawn  with  a  radius  =  AO3  intersect- 
ing the  sides  in  points  E,  F,  G,  etc. 
Show  that  the  figure  EFGHIJKL  is  a 
regular  octagon.      This  method  is  very 
useful  in  practice  as  a  quick  and  accur- 
ate construction. 

14.  Draw    a    square    side    of    which    is  3 
inches.     Construct    a    regular  octagon 
by  the  method  of  Exercise  13.      Calcu- 
late the  side  of  the  octagon. 


REGULAR   POLYGONS   AND    CIRCLES  257 

15.  Show  that  a  side  of  a  square  from  which  a  regular  octagon  of 
side  s,  may  be  cut  by  the  method  of  Exercise  13,  is  given  by 
the  formula,  s'  =  s(\/2  +  1).     Also  s  =  s'(\/2  —  1). 

16.  Find  the  side  of  a  square  from  which  a  regular  octagon  whose 
side  is  2  inches  may  be  cut. 

347.  Astronomical  Problems. — All  sections  of  the  earth 
by  a  plane  may  be  considered  (except  in  the  most  accurate 
work)  as  circles. 

EXERCISES 

1.  Taking  the  diameter  of  the  earth  as  41,780,000  feet,  calculate 
the  length  in  feet  of  one  minute  of  arc  on  the  equator  or  on  a 
meridian,  or  on  any  "great  circle." 

2.  The  earth  rotates  on  its  axis  in  24  hours.    Calculate  the  velocity 
of  a  point  on  the  equator  in  miles  per  hour. 

3.  The  earth  moves  about  the  sun  in  an  orbit  that  is  approximately 
circular  with  a  mean  radius  of  90,000,000  miles,  in  about  365 
days.     Calculate  its  velocity  in  the   orbit  in  miles  per  hour. 

The  Size  of  the  Earth. — The  most  accurate  method  of 
determining  the  size  of  the  earth  is  by  the  measurement  of 
an  arc  of  a  " great  circle"  on  the  earth's  surface.  A  and 
B  are  two  places  on  the  same  meridian;  AC  and  BD  are 
the  horizontal  planes  at  these  points.  Observations  are 
made  on  a  star  S  in  the  meridian  plane  of  arc  AB,  the  angles 
of  elevation,  CAS  and  DBS,  being  measured  simultaneously. 
Eratosthenes  made  a  fairly  accurate  calculation  of  the 
diameter  of  the  earth  by  this  method  (using  the  sun  as  the 
object  of  observation)  near  Alexandria,  Egypt,  about  250 
B.C.  S 


17 


258  PLANE    GEOMETRY 

EXERCISES 

1.  Show   that     Z.AOB  =   /.DBS  ~   /.CAS,    BS   and   AS   being 
parallel. 

2.  If    /.DBS  is  measured  67°  31',  and   /.CAS  is  measured  65°  43' 
find   /.AOB. 

3.  If  arc  AB  is  measured  124.5  miles,  calculate  the  circumference 
and  diameter  of  the  earth  in  miles. 

348.  THE  RADIAN. — This  is  a  new  unit  of  angle  and  arc. 
It  is  the  central  angle  of  a  circle  (or  its  arc)  whose  arc  is  the 
length  of  the  radius.     Do  not  confuse  this  with  the  central 
angle  subtended  by  a  chord  equal  to  the  radius.     The 
Radian  is  not  used  in  elementary  geometry,  but  is  an  im- 
portant unit  in  higher  mathematics. 

EXERCISES 

1.  Calculate  the  number  of  degrees  in  a  radian. 

Q£JA° 

Help. — Radian  =  -7, — 

2.  What  central  angle  is  subtended  by  a  chord  equal  to  the  radius? 

3.  How  many  radians  are  contained  in  an  arc  of  137°? 

349.  The  Circle  and  Regular  Polygon  in  Design. — Many 
ornamental  designs  for  rugs,  carpets,  wall  paper,  china  and 
glass-ware  decorations,  lace  and  embroidery,  tiling,  stone- 
tracery,  etc.,  are  based  upon  the  regular  polygon  and  its 
inscribed  and  circumscribed  circles.     Refer  to  a  ''Source 
Book  of  Problems  for  Geometry,"   by  Sykes;  Allyn  and 
Bacon;  for  an  analysis  of  numerous  designs  of  this  kind. 
See  also  195  and  321. 


EXERCISES 

1.  Lay  out  some  of  these  designs  on  squared  paper,  on  a  larger  scale. 

2.  Make  some  original  designs  based  upon  regular  polygons  and 
circles. 

3.  Sketch  some  designs  which  you  see  in  use.     Draw  the  principle 
lines  of  the  designs  and  omit  details  and  scrolls. 


REGULAR   POLYGONS   AND    CIRCLES  259 

FIELD  WORK 

350.  Design  and  lay  out  a  running  track  which  is  4,  6  or  8 
laps  to  the  mile.  Refer  to  345,  Fig.  1.  The  available 
rectangular  plat  should  first  be  measured  approximately 
and  laid  out  on  paper  to  scale.  The  shape  and  dimensions 
of  the  standard  line  is  then  determined  and  laid  out  on  the 
plat.  The  point  A  is  then  located  on  the  ground  from  a 
measurement  scaled  from  the  map;  AG  =  1J  feet.  Lay 
out  A B  parallel  to  the  side  of  the  field.  Erect  a  perpendi- 
cular to  AB  at  point  A,  and  set  stakes  at  points  0  and  0\. 
In  the  same  way  set  stakes  at  points  D  and  C.  The  per- 
pendiculars are  most  accurately  laid  out  with  a  transit,  but 
a  tape  may  be  used.  Measure  DC  as  a  check.  The  half 
circles  AMD  and  BNC  are  laid  out  by  holding  one  end  of  the 
tape  on  stakes  0  and  0\  and  carrying  the  end  around,  say, 
4  paces  at  a  time  and  driving  stakes  at  each  point  on  the 
curves'  thus  found.  Set  stakes  on  lines  AB  and  DC  about 
10  feet  apart.  The  outside  line  of  the  track  is  laid  out,  say, 
10  feet  from  the  inside  line.  Points  opposite  A,  B,  C,  D  are 
first  located.  The  arcs  may  be  laid  out  from  centers  0  and 
Oij  or  if  the  radius  is  too  great  for  the  length  of  tape,  stakes 
may  be  set  by  sighting  the  stakes  which  are  already  placed 
on  the  inner  arcs,  in  line  with  the  center  hub  stakes  0  and 
0,. 


CHAPTER  XVI 
AREAS 

MEASUREMENT  OF  AREAS.     PRINCIPLES  DETERMINED 
EXPERIMENTALLY 

351.  An  area  is  a  definite  part  of  a  surface.  A  new  unit 
is  required  for  the  measurement  of  an  area. 
The  unit  of  area  is  the  area,  or  surface,  or  extent, 
of  a  square,  each  side  of  which  is  a  unit  sect. 
Thus  unit  A  is  a  square  centimeter,  and  unit  B  is 
a  square  inch. 

Equivalent  areas  contain  the  same  area 
measure,  but  may  be  different  in  shape. 

A  segment  and  a  sector  of  a  circle  are 
certain  parts  of  the  area  of  a  circle. 
Refer  to  267. 

An  annulus  is  the  area  included  be- 
tween two  concentric  circles. 

Only  plane  areas  are  considered  in  this  text. 

Abbreviation. — Equivalent;  is  equivalent  to;  =c=. 

EXERCISES 

1.  What  geometric   units   have   been   previously   considered,   for 
measuring:  (a)  sects;  (6)  angles;  (c)  arcs?     Recall  their  defini- 
tions. 

2.  What  unit,  or  units,  are  used  to  measure  areas  the  size  of  this 
page?  the  area  of  a  room?  of  a  farm?  of  a  country? 

3.  Write  a  table  of  equivalents  between  different  area  units. 

4.  Are  equivalent  areas  necessarily  congruent? 

5.  Draw  a  square,  a  circle,  an  oblong  and  an  equilateral  triangle 
which  are  approximately  equivalent. 

6.  Add  geometrically,  a  square  2  inches  on  a  side  and  an  oblong 
2  inches  by  3  inches. 

260 


AREAS 


261 


7.  Subtract  geometrically,  the  areas  of  Exercise  6. 

8.  If  two  equal  rectangles  are  added  to  or  subtracted  from  two  equal 
triangles,    are   the   results   equivalent?     Are   they    necessarily 
congruent? 

9.  Draw  an  annulus  of  which  the  radii  of  the  bounding  circles  are 
1.2  and  1.7  inches,  respectively. 


352.  Experiment  I. — The  area  of  a  rectangle. 

Draw  rectangles;  (a)  2  inches 
by  3  inches,  full  size;  (6)  2| 
inches  by  3  inches;  (c)  2J  inches 
by  3J  inches.  Divide  into  square 
inches  and  fractions  thereof; 
count  the  area  measures  in 
units  and  fractions.  Observe 


the  relation  between  the  area  measure  of  each  rectangle 
and  the  sect  measures  of  length  and  width.  Squared 
paper  may  be  used  in  this  and  other  experiments  in  areas. 
State  result. 


EXERCISES 

1.  State  a  formula  for  the  area  R  of  a  rectangle,  in  terms  of  the 
width  w,  and  the  length  I. 

2.  Solve  the  formula  of  Exercise  1  for  w  and  for  I. 

3.  Draw  a  square  foot  to  one-half  linear  scale,  and  divide  it  into 
square  inches  to  one-half  linear  scale.     How  many  square  inches 
are  contained  in  the  square  foot? 

4.  Calculate  the  number  of  square  inches  (that  is,  the  area)  in  a 
rectangle  which  is  15  inches  by  8  inches. 

5.  How  many  square  feet  are  in  a  rectangle  18  inches  by  24  inches? 

6.  What  is  the  length  of  a  rectangle  whose  area  is  750  square  inches 
and  width  15  inches? 

7.  Draw   a   square    inch    A  BCD;   cut  it  A  £ B 

along   a   line   DE;   place   the  portion 

DBA,  cut  off,  so  that  AD  coincides 
with  BC.  Is  the  resulting  figure 
equivalent  to  the  original  square 
inch?  May  this  new  figure  also  be 
called  a  square  inch,  since  it  contains 
an  area  equal  to  that  of  the  given 
square  inch? 


262 


PLANE    GEOMETRY 


353.  Experiment  II. — The  area  of  an  oblique  parallelo- 
gram. 

Draw  rhomboids:  (a)  with  base  3  inches  and  altitude  2 
inches  (full  size);  (6)  with  base  3J  inches  and  altidude  2J 


inches.  Make  the  angle  x  equal  the  angle  x  of  352,  Exercise 
7.  Divide  the  rhomboids  into  units  and  fractions  thereof, 
of  the  shape  of  the  unit  referred  to  in  352,  Exercise  7. 
Observe  the  relation  between  the  area  measures  and  the 
dimensions  of  the  figures.  State  result. 

EXERCISES 

1.  State  a  formula  for  the  area  P  of  a  rhomboid  in  terms  of  the  base 
6,  and  the  altitude  a. 

2.  Solve  the  formula  of  Exercise  1  for  a  and  6. 

3.  Calculate  the   area  of  an  oblique  parallelogram  of  which  the 
base  is  25  feet  and  altitude  is  12  feet. 

4.  Perform    Experiment  II   by   cutting   the   rhomboid    on    some 
altitude,  and  placing  the  portion  cut  off  in  such  a  position 
as  to  form  a  rectangle.     Obtain  the  area  measure  of  this  rec- 
tangle, which  must  equal  that  of  the  given  parallelogram. 

5.  If  a  rhomboid  is  drawn  with  the  sides  including  any  other  angle 
than  the  angle  x  selected  for  the  figures  of  Experiment  II,  would 
the  result  be  the  same? 


354.  Experiment  III. — The  area  of  a  triangle. 
Draw  triangles  of  different  shapes.     Trace  on  thin  paper; 
revolve  the  traced  figures  on  the  points  A,B,C,  respectively, 


AREAS 


263 


which  are  the  middle  points  of  one  side  of  the  triangles,  and 
draw  the  symmetrical  impressions  of  the  triangles  as  indi- 
cated in  the  second  figure.  What  figure  is  formed  in  each 


case  by  two  adjoining  congruent  triangles?  What  is  the 
area  of  the  complete  figure?  What  is  the  area  of  the  single 
given  triangle?  State  result. 

EXERCISES 

1.  Write  a  formula  for  the  area  T,  of  a  triangle.     Solve  for  each 
letter. 

2.  Calculate  the  area  of  a  triangle  of  which  a  =  8",  b  =  10". 

3.  Calculate  the  base  and  altitude  of  a  triangle  whose  area  =  40 
square  inches,   and  of  which  the  base  exceeds  the  altitude  by 
2  inches. 

4.  Two  sides  of  a  triangle  are  20  and  36,  and  the  included  angle  is 
45°.     Calculate  the  area.     Help. — First  find  the  altitude. 

5.  Perform  Experiment  III  by  dividing  the  given  triangles  into 
square  units  and  fractions  thereof.     Use  squared  paper.     This 
method  employs  no  analysis  and  is  purely  experimental. 

355.  Experiment  IV. — The  area  of  a  trapezoid. 
Draw  trapezoids;  trace;  revolve  on  the  middle  point  of 
one  of  the  non-parallel  sides.     What  figure  is  formed  by  two 


adjoining  trapezoids?     What  is  the  area  of  this  figure? 
What  is  therefore  the  area  of  the  trapezoid? 


264 


PLANE  GEOMETRY 


EXERCISES 

1.  State  the  result  of  Experiment  IV  as  a  formula.     Solve  for  each 
letter. 

2.  The  dimensions  of  a  trapezoid  are  61  =  10,  bz  =  18,  a  =  8. 
Calculate  the  area,  S. 

3.  The  bases  of  a  trapezoid  are  15  and  25;  one  of  the  non-parallel 
sides  is  10;  and  its  projection  on  a  base  is  6.     Calculate  the  area. 

4.  In  a  trapezoid,  61  =  120,  62  =  190,  a  side  =  100,  the  angle  in- 
cluded between  the  100  side  and  the  190  base  =  60°.     Calcu- 
late the  area. 

5.  The  area  of  trapezoid  is  400  square  units,  the  altitude  is  30  units, 
one  base  is  10  units.     Find  the  other  base. 

6.  Perform  Experiment  IV  by  dividing  the  given  trapezoids  into 
square  units  and  fractions  thereof. 

356.  Experiment  V.— The  area  of  a  circle. 

Draw  a  circle  about  4  inches  in  diameter  and  cut  it  out. 
Cut  it  on  a  diameter  AB.  Cut  each  semicircle  into  the 
same  number  of  equal  sectors;  say  into  8  sectors.  Place 
the  sectors  as  shown  in  Fig.  2.  How  may  the  resulting 
figure  be  made  to  approximate  more  and  more  closely  to  a 
rectangle  as  in  Fig.  3?  What  are  the  base  and  altitude  of 
this  rectangle?  What  is  therefore  the  area  of  a  circle? 


246 


FIG.    2. 


FIG.  1. 


FIG.   3. 


EXERCISES 


1.  State  the  area  C  as  a  formula  in  terms  of  r  (or  d)  and  c. 

2.  State  the  value  of  C  in  terms  of  r  alone  (except  that  IT  will 
enter);  in  terms  of  d;  in  terms  of  c.     Refer  to  337   Corollary  2. 

3.  Solve  the  formulas  of  Exercise  2  for  r,  d  and  c,  respectively. 

4.  Calculate  the  area  of  a  circle  whose  radius  is  10". 

5.  Calculate  the  area  of  a  circle  whose  circumference  is  10". 


AREAS  265 

G.  Calculate  the  radius  of  a  circle  whose  area  is  7854  square  inches. 

7.  Perform  Experiment  V  by  drawing  a  circle  on  squared  paper 
and  counting  the  area  measure. 

8.  Show  by  a  method  similar  to  that  employed  in  Experiment  V, 
that  the  area  of  a  sector  equals  one-half  the  product  of  the  arc 
and  the  radius.     Express  this  as  a  formula. 

9.  Express   the   area  of   a  sector   in  terms  of  its 
central  angle,  n  degrees,  and  its  radius  r. 

Help. — Arc  a  =  =•=?.  2irr. 

OOU 

10.  Calculate  the  area  of  a  sector  whose  radius  is  10" 
and  of  which  the  length  of  arc  is  12". 

11.  Calculate  the  area  of  a  sector  whose  radius  is  20"  and  central 
angle  70°. 

357.  EXPERIMENT  VI. — THE  RATIO  OF  Two  AREAS. 
Draw  two  rectangles  having  different  bases  and  altitudes. 

Divide  them  into  square  inches  and  fractions 
thereof.  Find  the  area  measure  of  each  rec- 
tangle and  express  as  a  ratio. 

Divide  the  two  rectangles  into  any  other 
arbitrary  square  unit  s,  and  again  find  their  area  measure 
in  terms  of  this  unit,  and  express  as  a  ratio.  Observe 
how  the  ratios  compare.  State  result. 

EXERCISES 

1.  A's  farm  contains  twice  as  many  square  feet  as  B's  farm  con- 
tains.    Does  A's  farm  also  contain  twice  as  many  square  kilo- 
meters as  B's  farm  1 

2.  If  two  figures  are  drawn  to  some  part-size  scale,  will  the  ratio  of 
the  areas  of  the  figures  be  the  same  as  in  the  full-size  figures? 

358.  EXPERIMENT  VII. — THE  RATIO  OF  THE  AREAS  OF 
SIMILAR  FIGURES. 

(a)  Draw  a  rectangle.  Draw  other  similar  rectangles 
with  ratios  of  similitude  to  the  first  rectangle  of  2  :  1 ;  3  :  1 ; 
J£  :  1;  respectively:  that  is,  the  length  and  width  of  the 
second  rectangle  are  each  double  the  corresponding  dimen- 
sions of  the  first  rectangle;  etc.  Divide  these  rectangles 
into  smaller  rectangles  of  the  size  and  shape  of  the  first 
rectangle,  and  fractions  thereof.  Observe  how  the  ratios 


266  PLANE   GEOMETRY 

of  the  areas  compare  with  the  ratios  of  similitude  of  the 
figures. 

(6)  Draw  similar  triangles.  Divide  the  larger  triangles 
into  smaller  triangles  of  the  size  and  shape  of  the  smallest 
triangle  of  the  series.  With  this  triangle  as  a  unit  of 
measure,  observe  how  the  ratios  of  the  areas  of  the  triangles 
compare  with  their  ratios  of  similitude.  State  the  general 
result. 

EXERCISES 

1.  State  the  result  as  a  formula  in  terms  of  the  areas  S  and  S',  and 
a  ratio  of  similitude  —  or  ^7* 

2.  The  area  of  a  rectangle  is  5  square  inches.     What  is  the  area  of 
a  similar  rectangle  whose  ratio  of  similitude  to  the  first  is  10  :  1? 

3.  The  area  of  one  rectangle  is  36  times  the  area  of  a  similar  rec- 
tangle.    The  base  of  the  larger  rectangle  is  12  feet.     What  is 
the  base  of  the  smaller  rectangle? 

4.  The  area  of  an  equilateral  triangle  is  10  square  inches.     What 
is  the  side  of  an  equilateral  triangle  whose  area  is  40  square  inches, 
expressed  in  terms  of  the  side  s,  of  the  first  triangle? 

5.  Show  by  drawings  that  the  law  of  the  ratio  of  the  areas  of  two 
similar  figures  holds  for,  (a)  parallelograms,  (6)  trapezoids. 
Help. — In  the  figures  of  trapezoids,  a  cutting  and  re-uniting  of 
the  figure  is  necessary. 

369.  REVIEW  EXERCISES 

1.  Is  it  correct  to  say  that  a  number  of  inches  multiplied  by  a 
number  of  inches  results  in  a  number  of  square  inches?     Can 
two  denominate  numbers  be  multiplied?     If  a  denominate 
number  is  multiplied  by  an  abstract  number,  what  kind  of 
number  results? 

2.  Show  that  the  curved  (convex)  surface  of  a  circular  cylinder, 
bounded  by  parallel  planes  perpendicular  to  the  axis,  is  a  rec- 
tangle.    State  a  formula  for  the  area  of  the  convex  surface. 
For  the  total  area. 

3.  Show  that  the  convex  surface  of  a  right  circular  cone  is  a  sector. 
What  is  its  arc?     Its  radius?     State  a  formula  for  the  area  of 
the  convex  surface.     For  the  total  area. 

4.  The  circumference  of  a  tree  is  80  inches.     Find  the  area  of 
cross-section. 

5.  A  horse  is  tied  to  a  stake  by  a  rope  50  feet  long.     Over  what 
area  can  he  graze? 


AREAS 


267 


6.  A  horse  is  tied  to  a  stake  in  the  corner  of  a  square  court  by  a 
rope  50  feet  long.     Over  what  area  can  he  graze? 

7.  A  horse  is  tied  to  a  stake  at  the  corner  of  a  square  building. 
The  rope  is  100  feet  long,  and  the  building  is  50  feet  square. 
Over  what  area  can  he  graze? 

8.  How  many  square  feet  will  be  enclosed  by  a  rope  120  feet  long, 
if  it  is  laid  down  on  the  ground  (a)  in  the  form  of  a  square;  (6) 
in  the  form  of  an  equilateral  triangle;  (c)  in  the  form  of  a  circle? 

9.  What  may  we  conclude  is  the  greatest  (or  maximum)  area  for 
a  given  perimeter,  based  upon  the  results  of  Exercise  8? 

10.  What  part  of  a  circle  is  the  area  of  a  sector  of  110°?     Of  86°  32'? 

APPLICATIONS 

360.  Surveying. — Land  is  generally  valued  and  sold  in 
terms  of  its  area  measure.  City  lots  are  sometimes  rated 
in  terms  of  the  "front  foot." 


If  it  is  sold  for  $1200, 


EXERCISES 

1.  A  town  building  lot  is  120  feet  by  80  feet, 
what  price  is  obtained  per  square  foot? 

2.  A    farm    is    bounded    by    four 
straight  lines.      A  diagonal  AC 
is  560.75  feet;  the  distances  BE 
and  DF,  perpendicular  to  AC,  are 
respectively     93.44     feet     and 
612.20  feet.     Find  the  area  of 
the  farm  in  square  feet.     (The 
figure  is  not  drawn  to  scale.) 

361.  Estimating. — Estimates  for  painting,  roofing,  pav- 
ing, flooring,  wall  paper,  carpets,  etc.,  are  made  in  terms  of 
the  areas  involved. 


EXERCISES 

1.  Find  the  cost  of  laying  a  cement  pavement 
on  the  court  of  which  the  plan  is  given,  at 
$2.50  a  square  yard.     The  dimensions  are 
in  feet. 

2.  A  room  is  20  feet  long,  12  feet  wide,  and 
10  feet  high.     Find  the  cost  of  plastering 
the  walls  and  ceiling  of  the  room  at  20 
cents  a  square  foot,  deducting  25  percent 

of  the  area  for  doors,  windows  and  woodwork. 


40 


18 


268  PLANE    GEOMETRY 

3.  What  is  the  cost  of  laying  an  asphalt  pavement  per  mile  of  road, 
14  feet  wide,  at  $4.50  per  square  yard? 

4.  A  barn  roof  is  32  feet  wide  at  the  eaves,  12  feet 
high  to  the  ridge,  and  40  feet  long.     Allow  two 
feet  for  overhang  at  the  eaves.     Find  the  cost 
of  roofing  with  tar  paper  at  5  cents  a  square    | 
foot,  which  allows  for  overlapping  and  waste. 

5.  How  many  sheets  of  paper  8  by  11  inches,  can  be  cut  from  100 
sheets  24  by  44  inches? 

362.  PROBLEMS  FOR  FIELD  WORK. — (1)  Find  the  area 
covered  by  a  building  all  of  the  angles  of  which  are  right 
angles,  by  measuring  all  horizontal  dimensions.  Draw  a 
plan  and  divide  the  area  into  rectangles. 

2.  Find  the  area  covered  by  any  existing  or  proposed 
paths,  roads,  pavements,  etc.  The  cost  may  be  estimated 
by  procuring  a  trade  price  of  the  required  pavement  per 
square  foot  or  square  yard. 


CHAPTER  XVII 
AREAS 

CLASSIFICATION  AND  EXPLANATION  OF  PRINCIPLES 

363.  The  First  Step  in  Classification.    Area  Postulates.— 
(1)  If  an  area  is  divided  and  the  portions  are  placed  in  any 
other  relative  position  without  overlapping,  the  resulting 
area  is  equivalent  to  (that  is,  equal  in  measure  to)  the 
original  area. 

2.  The  unit  of  area  is  a  unit  of  the  same  kind,  and  is  a 
square  (or  its  equivalent)  each  side  of  which  is  a  unit  sect. 

3.  The  area  of  a  rectangle  is  the  unit  of  measure  multi- 
plied by  the  product  of  the  length  and  width  of  the  rec- 
tangle; or,  it  equals  the  product  of  the  length  and  width 
(or  base  and  altitude) . 

4.  The  area  of  a  circle  is  the  limit  which  the  areas  of 
inscribed  and  circumscribed  regular  polygons  approach,  as 
the  number  of  sides  is  indefinitely  increased. 

5.  The  areas  of  two  similar  rectangles  are  proportional 
to  the  squares  of  two  homologous  sides. 

THE  SECOND  STEP  IN  CLASSIFICATION 

364.  Theorem  I. — The  area  of  an  oblique  parallelogram 
equals  the  product  of  the  base  and  altitude. 


D         E  C 

Hypothesis. — An    oblique    parallelogram    A  BCD,    with 
base  DC  =  b,  and  altitude  AE  =  a. 
Conclusion. — Area  A  BCD  =  ab. 

269 


270  PLANE    GEOMETRY 

Analysis. 

STATEMENT  REASON 

1.  Extend  DC  and  draw    Postulate 
BF±DC 

2.  &ADE  ^  ABCF  Postulate:   2   sides   and   in- 

clude angle  =,  etc. 

3.  DC  =  EF  =  b  Axiom. 

4.  Area     ABFE  =c=  Area    363;  Postulate  1. 
ABCD 

5.  Area     ABFE   =  ab        363;  Postulate  3. 

6.  .'.  Area  ABCD  =  ab         Axiom. 

Corollaries. —  (1)  The  areas  of  two  parallelograms  which 
have  the  same  base  and  different  altitudes  are  proportional 
to  the  altitudes. 

2.  State  a  similar  corollary  for  parallelograms  which 
have  different  bases  and  the  same  altitude. 

EXERCISES 

1.  Are  all  rectangles  which  have  their  bases  and  altitudes  equal, 
each  to  each,  congruent? 

2.  Are  all  oblique  parallelograms  having  their  bases  and  altitudes 
equal,  each  to  each,  congruent?     Are  they  equivalent? 

3.  Write  a  formula  expressing  the  ratio  of  the  areas,  P  and  P',  of 
two  rectangles  or  two  oblique  parallelograms,  whose  bases  and 
altitudes  are  respectively  b  and  a,  and  6'  and  a'. 

4.  Write  a  formula  expressing  the  ratio  of  the  areas  of  any  two 
parallelograms  whose  bases  are  each  equal  to  6,  and  whose 
altitudes  are  respectively  a  and  a'. 

5.  Write  a  formula  similar  to  that  of  Exercise  4  if  the  parallelo- 
grams have  equal  altitudes  a,  and  bases  b  and  b'  respectively. 

6.  Write  a  formula  for  the  area  S,  of  a  square  whose  side  is  s. 

7.  Write  a  formula  for  the  side  s  of  a  square  whose  area  is  S. 

8.  The  base  and  altitude  of  two  parallelograms  are  respectively 
12"  and  36",  and  12"  and  18".     What  is  the  ratio  of  their  areas  ? 

9.  Find  the  area  of  a  rectangle  of  which  the  base  is  10  inches  and 
the  diagonal  is  12  inches. 

10.  The  sides  of  an  oblique  parallelogram  are  10"  and  15";  the 
projection  of  the  shorter  side  on  the  longer  side  is  6".  Calcu- 
late the  area.  Construct  the  parallelogram. 


AREAS  271 

11.  Show  that  the  area  of  the  oblique  parallelogram  of  Theorem  I 
equals  the  side  AD  multiplied  by  the  altitude  drawn  to  that 
side. 

12.  Prove  by  similar  triangles  that  the  product  ab  =  the  product 
of  AD  by  the  altitude  drawn  to  AD. 

365.  Theorem  II. — The  area  of  a  triangle  equals  one-half 
the  product  of  the  base  and  altitude. 


Helps.— (1)  Draw  BE\\AC  and  CE\\AB;  (2)  area  EEC  A 
=  ab;  (3)  /.  area  A  ABC  =  J  ab. 

Corollaries. — (1)  The  areas  of  two  triangles  which  have 
the  same  base  and  different  altitudes  are  proportional  to  the 
altitudes. 

2.  State  the  Corollary  which  corresponds  to  364,  Corol- 
lary 2. 

EXERCISES 

1.  Show  that  the  area  of  the  triangle  ABC  may  be  expressed  as  one- 
half  the  product  of  any  side  by  the  altitude  drawn  to  that  side. 

2.  Show  that  the  area  of  a  right  triangle  equals  one-half  the  product 
of  the  perpendicular  sides. 

3.  Write  formulas  expressing  the  ratios  between  the  areas  of  two 
triangles  T  and  T',  the  relations  between  the  dimensions  of 
which  correspond  to  those  stated  for  parallelograms  in  364, 
Exercises  3,  4  and  5. 

4.  Find  the  area  of  a  triangle  of  which  two  sides  are  12"  and  20", 
and  the  projection  of  the  shorter  side  on  the  longer  is  8". 

5.  The  mathematical  treatise  of  Ahmes  gives  the  area  of  an  isosceles 
triangle  whose  sides  are  10,  10,  4,  as  20.     What  formula  did 
Ahmes  employ?     What  is  the  correct  area?     When  is  Ahmes' 
formula  approximately  correct? 

6.  Find  the  area  of  a  triangle  of  which  two  sides  are  20  and  35  and 
the  included  angle  is  90°. 


272  PLANE    GEOMETRY 

7.  Find  the  area  of  a  triangle  of  which  two  sides  are  20  and  35  and 
the  included  angle  is  45°. 

8.  Find  the  area  of  a  triangle  of  which  two  sides  are  20  and  35  and 
the  included  angle  is  135°. 

366.  Theorem  III. — The  area  of  a  trapezoid  equals  one- 
half  the  product  of  the  sum  of  the  bases  by  the  altitude. 

h,  6z 


a 


i 
i 

i 


Write  out  the  analysis. 

EXERCISES 

1.  Find  the  area  of  a  trapezoid  whose  bases  are  20"  and  35"  and 
whose  altitude  is  15". 

2.  Find  the  other  base  of  a  trapezoid  of  which  the  area  is  750  square 
feet,  the  known  base  30  feet,  and  altitude  40  feet. 

3.  Find  the  altitude  of  a  trapezoid  of  which  the  area  is  1200  square 
feet,  and  the  bases  are  30  and  20  feet,  respectively. 

4.  The  method  used  by  the  ancient  Egyptians  for  finding  the  area 
of  a  piece  of  land  in  the  form  of  a  trapezoid  was  to  multiply 
half  the  sum  of  the  parallel  sides  by  one  of  the  other  sides. 
Is  this  method  ever  correct?     When  is  it  approximately  correct? 

THE  THIRD  STEP  IN  CLASSIFICATION 

367.  Theorem  IV. — The  area  of  a  triangle  can  be  expressed 
in  terms  of  the  three  sides. 

Helps. — Combine  the  result  of  Theorem  II,  365,  and  the 
value  of  an  altitude,  265. 
T  =\A(s  -  a)(s  -  b)(s^-  c). 

EXERCISES 

1.  Find  the  area  of  a  triangle  whose  sides  are  10,  12  and  14. 

2.  Find  the  area  of  the  isosceles  triangle  of  365,  Exercise  5,  by  the 
formula  of  this  section. 

3.  Find  the  area  of  a  right  triangle  whose  sides  are  30,  40  and  50 
by  the  formula  of  this  section.     Check  by  the  shorter  method 
of  366. 


AREAS 


273 


368.  Theorem  V. — The  area  of  a  regular  polygon  equals 
one  half  the  product  of  the  perimeter  by  the  apothem. 


Write  out  the  analysis. 

Corollary. — //  s  =  a  side,  n  =  the  number  of  sides,  a 

the  apothem,  then  P  =  —=-• 


EXERCISES 

1.  Find  the  area  of  a  regular  octagon  of  which  a  side  is  24.85  inches 
and  the  apothem  30  inches. 

2.  Find  the  area  of  a  regular  hexagon  whose  side  is  40. 

3.  Use  the  formula  of  Theorem  V  to  find  the  area  of  a  square  whose 
side  is  40.     Check  by  the  formula  S  =  ab  (363,  Postulate  3). 

369.  Theorem  VI. — The  area  of  any  circumscribed  polygon 
equals  one-half  the  product  of  the  perimeter  by  the  radius  of  the 
inscribed  circle. 


Write  out  the  analysis. 

18 


274 


PLANE    GEOMETRY 


370.  Theorem  VII. — The  area  of  a  circle  equals  one-half 
the  product  of  the  circumference  and  radius. 

D 


Hypothesis. — Circle  0,  with  radius  r,  and  circumference  c. 

TC 

Conclusion. — Area  of  circle,  S  =  -*• 

Analysis. 

STATEMENT 

1.  Circumscribe  any  regular  poly- 
gon A  BCD  .   .   .  ,  about  circle 
0,  with  an  area  P,  and  a  peri- 
meter p. 


REASON 


2.P-f 

3.  Indefinitely  increase  the  num- 
ber of  sides  of  polygon  A  BCD .'. 

4.  Area  P  =  Area  S  ' 

5.  And  p  =  c 

.   rp      re 
'   2        2 

7.  But  P  always  =  ~,  and 

•  s  -rc 

^ 

Corollaries. 

1.  S  =  irr2  or  3.1416r2. 

2.  S  =  ~  or  .7854d2. 


369;  Theorem  VI. 

363;  Postulate  4 
335;  Postulate  4  (c) 

335 ;  Limit  Axiom  3 
335;  Limit  Axiom  2 


3.  S  =  -A-  or  .0796c2. 
4?r 


AREAS  275 

4.  The  areas  of  two  circles  are  proportional  to  the  squares 
of  their  radii,  diameters  or  circumferences. 

EXERCISES 

o 

1.  The  formula  given  by  Ahmes  for  the  area  of  a  circle  is  S  =  ^.dz. 

Show  that  this  is  equivalent  to  a  value  of  TT  =  3.16. 

2.  Is  it  probable  from  the  evidence  of  the  formula  itself  that  Ahmes 
arrived  at  this  measure  by  comparing  areas  experimentally,  or 
by  deductive  reasoning? 

3.  Calculate  the  radius,  diameter,  and  circumference  of  a  circle 
whose  area  is  78.54  square  inches;  of  a  circle  whose  area  is  1200 
square  inches. 

4.  Find  the  area  of  a  sector  of  10°  in  a  circle  of  radius  30. 

5.  Find  the  area  of  a  sector  whose  arc  equals  twice  the  radius,  in  a 
circle  of  12"  radius. 

371.  Theorem  VIII. — The  radius  of  the  inscribed  circle 
of  a  triangle  may  be  expressed  in  terms  of  the  three  sides. 

Hypothesis. — AABC;  the  inscribed  circle  with  center  0 
and  radius  r. 

Conclusion. — r  may  be  expressed  in  terms  of  the  sides  a, 
b,  c,  of  the  triangle.  • 

Analysis.— (1)  Draw  OA,  OB,  OC;  (2)  Area  ABOC  = 

^,  etc.;  (3)  Area  A  ABC  =  q  +  k  +  c.  r  =  s  -  r;   (4)   But 


area  A  ABC  =  \/s(s  -  a)  (s  -  b)  (s  -  c);     (5)    .'.     r    = 
(s  —  a)  (s  —  b)  (s  —  c) 
s 

EXERCISES 

1.  Calculate  the  radius  of  the  inscribed  circle  of  a  triangle  whose 
sides  are  3,  4  and  5. 

2.  Calculate  the  area  of  the  triangle  of  Exercise  1,  using  the  formula 
S  =  sr;  and  the  value  of  r  found  in  Exercise  1. 

ADDITIONAL  THEOREMS 

372.  (1)  Two  triangles  of  which  two  sides  are  equal,  each 
to  each,  and  the  included  angles  supplementary,  are  equivalent; 
that  is,  equal  in  area. 


276  PLANE    GEOMETRY 

(2)  A    median    divides    a    triangle    into  two  equivalent 
triangles. 

(3)  The  area  of  a  trapezoid  equals  the  product  of  the 
altitude  by  the  median. 

(4)  The  area  of  any  quadrilateral  equals  one-half  the 
product  of  a  diagonal  by  the  sum  of  the  perpendiculars  drawn 
to  the  diagonal  from  the  two  opposite  vertices.     Write  the 
formula. 

(5)  //  the  diagonals  of  a  quadrilateral  are  perpendicular, 

the  area  is  S  =  — „-,  where  di  and  dz  are  the  diagonals. 

(6)  The  area  of  a  rhombus  equals  one-half  the  product  of 
the  diagonals. 

(7)  The  area  of  a  square  is  S  =  «  /  where  d  is  the  diagonal. 

(8)  The  diagonal  of  a  square  may  be  expressed  in  terms  of 
the  area. 

(9)  The  area  of  a  circumscribed  square  is  double  that  of  the 
inscribed  square  of  the  same  circle. 

(10)  The  area  of  an   equilateral  triangle  is  T  = — j— / 

where  s  =  a  side  of  the  triangle. 

Helps. — (a)  Derive  the  formula  from  the  area  formula, 

ab 
T  =-%,  (365) ;  (6)  Derive  it  from  T  =  \/s(s-a)(s-  6)(s-c), 

(367),  where  s  =  one-half  the  sum  of  the  sides  of  the  triangle. 

(11)  The  side  of  an  equilateral  triangle  may  be  expressed 
in  terms  of  the  area. 

Help.-s  =?VfV3 
o 

(12)  The  area  of  a  regular  hexagon  is  H  = — v~*i  where  s 

is  a  side  of  the  hexagon. 

(13)  The  side  of  a  regular  hexagon  may  be  expressed  in 
terms  of  the  area. 

(14)  The  area  of  an  annulus  may  be  expressed  in  terms  of 
the  radii,  or  of  the  diameters,  of  the  bounding  circles. 


AREAS  277 

Helps.—  (1)    S  =  Si  -  S2;    (2)  Si  =  irrS  and  S2  =  7rr22; 
(3)  Subtract  and  factor;  S  =  7r(ri  +  r2)(ri  —  r2). 

(15)  The  area  of  a  triangle  may  be  expressed  in  terms  of  the 
three  sides  and  the  radius  of  the  circumscribed  circle. 

Helps.—  Refer  to  308  and  367. 
_obc 
"  4R 

(16)  The  radius  of  the  circumscribed  circle  of  a  triangle 
may  be  expressed  in  terms  of  the  area  and  sides  of  the  triangle. 

(17)  The  area  of  a  minor  segment  of  a  circle  is  S  =  ™- 


irr2  —  ~;  where  x  is  the  central  angle  of  the  segment,  r  the 

z 

radius  of  the  circle,  c  the  length  of  the  chord  of  the  segment,  and 
p  the  perpendicular  upon  the  chord  from  the  center  of  the 
circle. 

Help.  —  Area    of    segment  =  area    of    sector  —  area    of 
triangle. 

(18)  State  the  formula  for  the  area  of  a  major  segment  of  a 
circle. 

(19)  The  difference  of  the  areas  of  a  circle  and  the  circum- 
scribed square  may  be  expressed  in  terms  of  the  radius  or 
diameter  of  the  circle. 

(20)  The  different  of  the  areas  of  a  circle  and  the  inscribed 
square  may  be  expressed  in  terms  of  the  radius  or  diameter  of 
the  circle. 

373.  EXERCISES 

1.  If  the  area  of  a  triangle  of  which  two  sides  and  the  included 
angle  are  respectively  37",  51"  and  65°,  is  approximately  216 
square  inches.     What  is  the  area  of  a  triangle  of  which  two 
sides  and  the  included  angle  are  respectively  37",  51",  and  115°? 

2.  Check  the  value  given  in  Exercise  1  by  constructing  one  of  the 
triangles  and  measuring  an  altitude. 

3.  A  diagonal  of  a  quadrilateral  is  27"  and  the  perpendiculars 
drawn  to  this  diagonal  from  the  other  two  vertices  are  respec- 
tively 10"  and  45".     Calculate  the  area  of  the  quadrilateral. 

4.  The  diagonals  of  a  rhombus  are  12"  and  20".     Calculate  the 
area. 


278  PLANE    GEOMETRY 

5.  Find  the  diagonal  of  a  square  whose  area  is  3200  square  inches. 

6.  The  area  of  a  square  circumscribed  about  a  circle  is  240  square 
inches.     Find  the  area  of  a  square  inscribed  in  the  same  circle. 

7.  Find  the  area  of  an  equilateral  triangle  whose  side  is  20  inches, 
using  the  formula  of  372,  Theorem  (10.) 

8.  The  area  of  an  equilateral  triangle  is  100  square  inches.     Find 
the  side  of  the  triangle. 

9.  Find  the  area  of  a  regular  hexagon  inscribed  in  a  circle  whose 
radius  is  12  inches. 

10.  Find  the  area  of  an  annulus  whose  radii  are  10"  and  20" 
respectively. 

11.  The  radius  of  one  of  the  bounding  circles  of  an  annulus  is  24 
inches  and  the  area  of  the  annulus  is  1200  square  inches.     Find 
the  radius  of  the  other  bounding  circle  (a)  if  the  given  circle  is 
the  smaller  of  the  two  circles;  and  (b)  if  the  given  circle  is  the 
larger  of  the  two  circles. 

12.  Find  the  area  of  the  triangle  of  308,  Exercise  1,  using  the  for- 
mula of  372,  Theorem  (15.) 

13.  Find  the  area  of  a  segment  of  120°  in  a  circle  whose  radius  is 
12  inches. 

14.  Find  the  difference  in  the  areas  of  a  circle  whose  radius  is  20 
inches,  and  its  inscribed  square,  using  the  formula  of  372. 
Theorem  (20.) 

THE  FOURTH  STEP  IN  CLASSIFICATION 

374.  EXERCISES 

1.  Construct  the  triangle  of  367,  Exercise  1,  to  a  smaller  scale. 
Measure    an    altitude.     Calculate   the    area   by   the   formula 
S  =  %ab  (365).     Compare  the  result  with  the  value  obtained  in 
the  Exercise  referred  to. 

Exercises  2  and  3,  367,  also  are  checks  on  the  formula  of  367. 

2.  Construct  the  triangle  of  371,  Exercise  1,  and  the  inscribed  circle. 
Measure  the  radius.     Compare  with  the  value  calculated  in 
the  Exercise  referred  to. 

3.  Construct  the  inscribed  and  circumscribed  squares  of  the  same 
circle.     Measure  a  side  of  each,  and  calculate  the  areas.     Does 
the  result  agree  with  the  principle  of  372,  Theorem  (9)  ? 

4.  Construct  an  equilateral  triangle  whose  side  is  20.     Draw  and 
measure  an  altitude  and  calculate  the  area  by  the  formula  of 
365.     Does  the  result  agree  with  the  value  obtained  in  373, 
Exercise  7? 

5.  Construct  a  triangle  whose  sides  are  10,  12  and  14  (Exercise  1). 
Construct  the  circumscribed  circle.     Measure  the  radius.     Cal- 


AREAS 


279 


culate  the  area  by  the  formula  of  372,  Theorem  (15).  Does  the 
result  agree  with  the  values  obtained  in  367,  Exercise  1,  and  in 
Exercise  1  of  this  section? 

RELATIONS  BETWEEN  AREAS 
THE  SECOND  STEP  IN  CLASSIFICATION 

375.  Theorem  IX. — The  areas  of  two  similar  triangles  are 
proportional  to  the  squares  of  any  two  homologous  sides;  or, 
to  the  square  of  their  ratio  of  similitude. 

A 

A' 


C' 


Hypothesis. — Similar  triangles  ABC  and  A'B'C'. 
Area  AABC         a2 


Conclusion. — 


Area  AA'B'C'      a'2 


Analysis. 

STATEMENT 

1.  Area  AABC  =  ~* 

2.  Area  AA'B'C'  =  * 

Area  AABC 


3. 


' 


5. 


2 

ha-a 


Area  AA'B'C'      ha'-a' 
ha    _  a 
17  ~  a' 

Area  AABC      _  a2 
Area  AA'B'C'  ~  a/2 


REASON 
Theorem 

Theorem 
Axiom 
Theorem  of 
Axiom 


Corollaries. — (l)The  areas  of  two  similar  triangles  are 
proportional  to  the  squares  of  homologous  altitudes,  medians, 
or  of  any  two  homologous  lines. 

2.  Any  homologous  lines  (sides,  altitudes,  etc.)  of  two 
similar  triangles  are  proportional  to  the  square  roots  of  the 
areas  of  the  triangles. 


280 


PLANE    GEOMETRY 


EXERCISES 

1.  The  sides  of  two  triangles  are  respectively  7,  10  and  14,  and  10  5, 
15  and  21.     What  is  the  ratio  of  the  areas  of  the  triangles? 

2.  The  area  of  a  triangle  is  four  times  the  area  of  a  similar  triangle. 
What  is  the  ratio  of  two  homologous  sides  of  the  triangles? 

THE  THIRD  STEP  IN  CLASSIFICATION 
376.  Theorem  X. — The  areas  of  two  similar  polygons  are 
proportional  to  the  squares  of  two  homologous  sides. 


Hypothesis. — Similar  polygons  X  and  Y. 
!r  =  T\  =  etc. 


Conclusion. —  - 


Analysis. 


STATEMENT 


REASON 


1.  Divide  -X"  and  Y  into  triangles 
similarly   placed,    A\,    B\  .   .   . 


2. 


5.  ±1   =  *i  = 


tri- 


Theorem  of 
angles. 

375;  Theorem  IX. 


6. 


A, 

A, 


Bl 


a 


A2 


7      ^  —   a*2  _   ^l2 

F  :=  a?  =  S2  = 
State  corollaries  similar  to  those  of  375. 


AREAS 


281 


EXERCISES 

1.  The  area  of  a  pentagon  is  50  square  inches;  the  ratio  of  similitude 
of  a  second  pentagon  similar  to  the  first  pentagon,  is  9:1.     Find 
the  area  of  the  second  pentagon. 

2.  The  areas  of  two  similar  quadrilaterals  are  50  and  100.     A  side 
of  the  first  quadrilateral  is  10;  what  is  the  homologous  side  of  the 
second  quadrilateral? 

377.  THEOREM  XI.  —  The  areas  of  two  triangles  which  have 
an  angle  of  one  triangle  equal  to  an  angle  of  the  other,  are 
proportional  to  the  products  of  the  sides  including  the  equal 
angles. 


c' 


Hypothesis.—  A  ABC  and  DEF  with 

Area  AABC  AB-AC 
C°ncluSWn'-  Are*  ADEF  -=D' 
Analysis. 

STATEMENT 

1.  Place    A  ABC   in    the    position 
B'DC'  and  draw  EC'. 
Area  ADB'C'      DB'      AB 
=  DE  =  DE 
DF       DF 


REASON 


0 
2' 


3' 


4' 


Area 

Area  ADEF 


Area 

Area  A  ABC 


AB  -  AC 


OCK    ^      .. 
365;  C°r°llary 
oc_    ^      „ 
365;  C°r°llary 

Axl°m 


Area  A  DE-  DF 

Corollary.  —  The  areas  of  two  triangles  which  have  an 
angle  of  one  supplementary  to  an  angle  of  the  other,  are 
proportional,  etc. 

EXERCISES 

1.  The  area  of  a  triangle,  two  of  whose  sides  are  10  and  12  and  of 
which  the  included  angle  is  60°,  is  103.92.  What  is  the  area  of 
a  triangle,  two  of  whose  sides  are  5  and  30  and  of  which  the  in- 
cluded angle  is  60°? 


282 


PLANE   GEOMETRY 


2.  What  is  the  area  of  a  triangle,  two  of  whose  sides  are  15  and  25 
and  of  which  the  included  angle  is  120°? 

378.  Theorem  XII. — The  area  of  the  square  drawn  on  the 
hypotenuse  of  a  right  triangle  equals  the  sum  of  the  areas  of 
the  squares  drawn  on  the  two  perpendicular  sides. 

6 


K  H 

State  the  hypothesis  and  the  conclusion. 

Helps.— (I)  Rectangle  CDEF  =c=  2  times  &BCD;  (2) 
Square  Z  ~  2  times  AACH;  (3)  ABCD  ±  AACH;  (4) 
.'.  Rectangle  CDEF  =c=  square  Z;  etc. 

This  theorem  has  been  analyzed  before  under  similar  tri- 
angles, but  there  is  an  essential  difference  in  the  method 
employed  here  and  that  of  the  former  analysis.  The  analy- 
sis here  given  was  employed  by  Euclid. 

More  than  three  hundred  different  analyses  of  the  theo- 
rem have  been  discovered. 

THE  FOURTH  STEP  IN  CLASSIFICATION 

379.  EXERCISES 

1.  Draw  an  irregular  hexagon;  draw  a  second  hexagon  similar  to 
the  first,  with  the  ratio  of  similitude  1  :  2.  Divide  the  hexagons 
into  triangles  in  any  way,  and  measure  a  side  and  an  altitude  to 
that  side  in  each  triangle.  Calculate  the  areas  of  each  hexagon 
by  adding  the  areas  of  the  triangles  into  which  it  is  divided. 
How  does  the  ratio  of  the  areas  of  the  hexagons  compare  with 
their  ratio  of  similitude?  Does  this  agree  with  the  Theorem 
of  376? 


AREAS  283 

2.  Draw  any  oblique  scalene  triangle;  draw  a  second  triangle  with 
an  angle  equal  to  an  angle  of  the  first  triangle.  Measure  a  side 
and  an  altitude  to  that  side  in  each  triangle  and  find  the  areas 
by  the  formula  of  365.  Find  also  for  each  triangle  the  products 
of  the  sides  including  the  equal  angles.  Compare  the  ratio  of 
the  areas  with  the  ratio  of  these  products,  by  reducing  each 
ratio  to  a  decimal.  Does  the  result  agree  with  the  principle  of 
377? 

ADDITIONAL  THEOREMS 

380. — (1)  The  square  on  the  base  of  a  right  isosceles  tri- 
angle is  equivalent  to  four  times  the  triangle. 

(2)  The  four  triangles  into  which  the  diagonals  divide  a 
parallelogram  are  equivalent. 

(3)  Straight  lines  drawn  from  the  middle  point  of  a  diag- 
onal of  a  quadrilateral  to  opposite  vertices  divide  it  into  two 
equivalent  parts. 

(4)  The  parallelogram  formed  by  joining  the  middle  points 
of  each  two  adjacent  sides  of  a  quadrilateral  is  equivalent  to 
one-half  the  quadrilateral. 

(5)  A  line  drawn  through  the  middle  point  of  the  median 
of  a  trapezoid,  cutting  the  bases,  divides  the  trapezoid  into 
two  equivalent  parts. 

(6)  The  triangle  formed  by  joining  the  middle  point  of  one 
of  the  non-parallel  sides  of  a  trapezoid  to  the  extremities  of 
the  other  non-parallel  side  is  equivalent  to  one-half  the  trape- 
zoid. 

(7)  The  sects  joining  the  middle  points  of  the  three  sides 
of  a  triangle  divide  it  into  four  equivalent  (congruent)  triangles. 

(8)  The  areas  of  two  parallelograms  having  an  angle  of  one 
equal  to  an  angle  of  the  other  are  proportional  to  the  products 
of  the  sides  including  the  equal  angles. 

(9)  The  sum  of  the  areas  of  the  squares  on  the  sects  drawn 
from   any  point  within  a  rectangle  to  two  opposite  vertices 
equals  the  sum  of  the  areas  of  the  squares  on  the  sects  drawn 
from  the  point  to  the  other  two  opposite  vertices. 

Helps. — Draw  lines  through  the  point  parallel  to  the  sides 
of  the  rectangle,  forming  right  triangles. 


284 


PLANE    GEOMETRY 


(10)  The  area  of  the  equilateral  triangle  drawn  on  the 
hypotenuse  of  a  right  triangle  equals  the 
sum  of  the  areas  of  the  equilateral  triangles 
drawn  on  the  other  two  sides. 


Helps. -(1)       =       ; 


Z 

X 


0* 

b*' 


Y+Z      a2  +  c2 


=  1;  (3)  :.X  =  Y  +Z. 


X  b 

(11)  State    and    analyze    a    general 
theorem    for    similar     polygons,     sug- 
gested by  Theorem  (10). 

(12)  The  area  of  a  triangle  of  which  one  angle  is  30°  equals 
one  quarter  the  product  of  the  including  sides.     Express  as  a 
formula. 

(13)  The  area  of  a  triangle  of  which  one  angle  is  45°  equals 
one  quarter  the  product  of  the  including  sides  times  the  square 
root  of  two.     Express  as  a  formula. 

(14)  Find  the  formula  for  the  area  of  a  triangle  of  which 
one  angle  is  60°  and  the  including  sides  are  b  and  c. 

(15)  Write  the  formulas  for  the  areas  of  triangles  of 
which  two  sides  are  b  and  c  and  the  included  angle  is  150°, 
135°  or  120°. 

(16)  //  semicircles  are  drawn  on  the  three  sides  of  a  right 
triangle,  as  in  the  figure,  the  sum  of  the 

crescents  X  and  Y  equals  the  area  T  of 
the  given  triangle. 

Helps. — (1)  Write  the  formulas  for 
the  semicircles  in  terms  of  their  diame- 
ters: (2)  write  the  sum  of  the  areas  X 
and  Y  as  an  algebraic  sum  of  T  and  the  semicircles. 

(17)  //  lines  are  drawn  from  any  point  within  a  parallelo- 
gram to  the  four  vertices,  the  sum  of  the 

areas  of  two  opposite  triangles  thus  formed 
is  equal  to  the  sum  of  the  areas  of  the  other 
two  opposite  triangles. 

(18)  A  parallelogram  is  divided  into  two 
equivalent  areas  by  a  line  drawn  through 
the  point  of  intersection  of  the  diagonals. 


AREAS  285 

(19)  A  trapezoid  is  divided  into  two  equivalent  areas  by  a 
line  which  joins  the  middle  points  of  the  bases. 

(20)  The  area  of  a  trapezoid  equals  the  product  of  one  of  the 
non-parallel  sides  by  the  perpendicular  drawn  to  that  side 
from  the  middle  point  of  the  other  of  the  non-parallel  sides. 

(21)  Lines  which  join  the  middle  points  of  the  sides  of  a 
parallelogram,  in  order,  form  four  equivalent  triangles,  and 
the  sum  of  the  areas  of  these  triangles  equals  the  area  of  the 
parallelogram  formed  by  the  four  lines. 

(22)  Lines  which  are  drawn  through  any  point  on  a  diagonal 
of  a  parallelogram,  parallel  to  the  sides  of  the  parallelogram, 
form  with  the  sides  of  the  given  parallelogram  four  smaller 
parallelograms,  two  of  which  are  equivalent. 

381.  REVIEW  EXERCISES 

1.  Find  the  non-parallel  sides  and  the  diagonals  of  an  isosceles 
trapezoid  of  which  the  area  is  144  square  units  and  the  altitude 
is  8  units  and  one  base  is  12  units. 

2.  Two  sides  of  a  triangle  are  12"  and  15"  and  the  altitude  to  the 
longer  of  the  two  sides  is  10".     Find  the  altitude  to  the  other 
of  the  two  sides. 

3.  The  perpendicular  sides  of  a  right  triangle  are  4  and  6.     Find 
the  area  of  the  triangle. 

4.  The  hypotenuse  of  a  right  triangle  is  20  and  its  base  is  12. 
Find  the  area  and  the  altitude  to  the  hypotenuse. 

5.  Find  the  side  of  a  rectangle  of  which  the  area  is  120  square 
inches  and  the  other  side  is  6  inches. 

6.  Find  the  diagonal  of  a  rectangle  of  which  the  area  is  48  square 
inches  and  base  8  inches. 

7.  The  base  and  altitude  of  a  rhomboid  are  12"  and  10"  respec- 
tively.    Find  the  area. 

8.  Two  sides  of  a  rhomboid  are  8"  and  14"  and  the  altitude  to  the 
longer  side  is  7".     Find  the  altitude  to  the  shorter  side. 

9.  Two  sides  of  a  parallelogram  are  50"  and  30"  and  the  included 
angle  is  3010.     Find  the  area. 

10.  Calculate  the  area  of  a  quadrilateral  of  which  one  diagonal  is 
120,  and  the  perpendiculars  upon  this  diagonal  from  two  oppo- 
site vertices  are  50  and  75. 

11.  Two  perpendicular  diagonals  of  a  trapezium  are  81  and  44. 
Calculate  the  area. 


286  PLANE   GEOMETRY 

12.  The  side  of  a  rhombus  is  10  and  a  diagonal  is  16.     Calculate 
the  other  diagonal  and  the  area. 

13.  Calculate  the  area  of  a  square  whose  diagonal  is  24.     Check 
by  finding  the  area  from  a  side. 

14.  Calculate  the  areas  of  the  circumscribed  and  inscribed  squares 
of  a  circle  whose  diameter  is  8. 

15.  The  area  of  a  square  is  72;  calculate  the  diagonal  without 
finding    the    side.       Check    by    finding   the  side   and   then 
the  diagonal. 

16.  The  side  of  an  equilateral  triangle  is  12.     Calculate  the  area. 
Check  by  first  finding  the  altitude. 

17.  Calculate  the  radius  of  the  circle  in  which  a  regular  inscribed 
hexagon  has  an  area  180. 

18.  The  difference  of  the  areas  of  a  circle  and  the  circumscribed 
square  is  85.84  square  inches.     Calculate  the  radius  of  the 
circle.     Check  by  finding  each  area  from  the  radius. 

19.  The  sum  of  the  four  segments  included  between  a  circle  and  the 
inscribed  square  is  456.64  square  inches.     Calculate  the  radius 
of  the  circle. 

20.  The  difference  between  the  areas  of  a  circle  and  the  regular 
inscribed  hexagon  is  1000  square  inches.     Calculate  the  radius 
of  the  circle  and  its  area. 

21.  The  area  of  a  circumscribed  square  is  25  square  inches.     What 
is  the  area  of  the  square  inscribed  in  the  same  circle? 

22.  Find  the  area  of  the  bases  of  a  cylinder  which  is  6  inches  long 
and  2  inches  in  diameter. 

23.  How  many  square  inches  are  in  the  convex  surface  of  a  cylinder 
12  inches  long  and  10  inches  diameter? 

24.  A  cylindrical  boiler  is  10  feet  long  and  4  feet  diameter.     How 
many  square  feet  of  sheet  metal  are  required  to  construct  its 
convex  surface? 

25.  Find  the  area  of  the  base  of  a  cone  which  is  6  inches  in  altitude 
and  2  inches  in  diameter  at  the  base. 

26.  Calculate  the  total  area  of  the  surface  of  a  cone  16  inches  alti- 
tude and  24  inches  diameter  of  base. 

Help. — Calculate    the    slant    height;    convex    surface;    area 
of  base. 

27.  The  sides  of  a  rhombus  and  the  shorter  diagonal  are  each  10". 
Calculate  the  longer  diagonal  and  the  area. 

28.  Calculate    and    arrange   in    order    of  length,   the   circumfer- 
ence of  a  circle,  the  perimeters  of  a  square,  an  equilateral 
triangle,   a  regular  hexagon;  all  of  the  same  area.     Let  the 
area  =  1.     In  which  figure  is  the  ratio  of  perimeter  to  area 
a  minimum? 


AREAS 


287 


29.  Show  by  a  figure  that  (a  +  6)2  =  a2  +  2ab  +  62. 
Help. — Show  the  equivalence  of  the  areas. 

30.  Show  by  a  figure  that   (a  -  6)2   =  a2  - 
2ab  +  62. 

31.  Show  by  a  figure  that  (a  +  6)  (a  —  6)  = 
a2  -  62. 

32.  Show  by  a  figure  that  (a  +  6)  (a  +  c)  = 
a2  +  a&  -f  ac  +  be. 

33.  Show  by  a  figure  that  (a  —  6)  (a  —  c)  =  a2  —  db  —  ac  +  be. 

34.  Show  by  a  figure  that  (a  +  b  +  c)2  =  a2  +  b2  +  c2  +  2a6  + 
2ac  +  26c. 

Euclid  gave  the  geometric  construction  for  Exercises  29  to  31  in 
the  "Elements."  Algebra  was  almost  unknown  at  that  time, 
but  many  familiar  algebraic  relationships  were  discovered  in 
geometric  form. 

35.  If  two  polygons  are  similar,  and  one  is  four  times  the  other 
in  area,  what  is  their  ratio  of  similitude?     What  if  one  is  double 
the  other? 

36.  Divide  the  altitude  of  a  triangle  into 
four  equal  parts  by  lines  parallel  to 
the  base.     What  parts  of  the  entire 
triangle  are  the  areas  of  the  strips? 
Help. — Calculate  the  areas  of  &ADE, 
AFG,  etc.,  in  terms  of  A  ABC. 

37.  The  homologous  sides  of  two  similar  polygons  are  5  and  10. 
Find  the  side  of  a  similar  polygon  equivalent  to  their  sum;  to 
their  difference. 

38.  The  area  of  an  equilateral  triangle  is  1200  square  inches.     Calcu- 
late the  side. 

39.  The  cross-section  of  a  cut  through  a  hill  is  a  trapezoid,  20  feet 


wide  at  the  bottom,  10  feet  deep,  the  angles  of  slope  x,  45°. 
Calculate  the  area  of  cross-section. 

40.  Write  a  formula  for  the  cross-section  of  Exercise  39,  when  the 
cut  is  w  feet  wide  at  bottom,  c  feet  deep,  slope  45°. 

41.  The  scale  of  a  map  is  100  miles  to  the  inch.     What  area  is  in- 
closed by  a  square  inch?     What  area  is  inclosed  by  a  rectangle 
\  inch  by  1£  inches? 

42.  If  a  rectangular  area  2  by  3  inches  is  known  to  represent  on  a 
certain  map  an  area  of  15,000  square  miles,  what  is  the  linear 
scale  of  the  map? 


288 


PLANE    GEOMETRY 


8     C 


43.  The  following  is  found  in  many  popular  lists  of  "trick"  prob- 
lems.    Draw  a  square  8  inches  on  a  side;  cut  it  as  shown; 
re-assemble  the  parts  to  form  a 

rectangle  5  by  13  inches.  How 
many  square  inches  are  ap- 
parently gained  by  the  opera- 
tion? If  this  were  true  what 
axiom  of  areas  would  be  con- 
tradicted? Show  the  fallacy  of  the  apparent  result  by 
calculating  the  length  of  DE  if  AC  and  BC  are  straight  lines. 
If  DE  =  3  inches,  can  both  these  lines  be  straight? 

44.  The  base,  6,  of  a  triangle  is  increased  by  an  amount  b' ;  how 
much  must  the  altitude  a  be  decreased  in  order  that  the  area 
shall  remain  unchanged? 

45.  The  altitude  and  base  of  a  triangle  are  a  and  b;  the  altitude  and 
base  of  a  similar  triangle  are  a'  and  b'.     Calculate  the  altitude 
and  base  of  a<  similar  triangle  equal  to  their  sum. 


Helps. 


+  a'6') 


+  r2;  b"  =         (ab  +  a'6') 


=  b\/l  +  r2;  where  a"  and  6"  are  the  required  base  and  alti- 
tude, and  r  is  the  ratio  of  similitude  of  the  given  triangles. 

46.  The  altitude  and  base  of  a  triangle  are  10  and  13;  the  altitude 
and  base  of  a  similar  triangle  are  20  and  26.     Calculate  the 
altitude  and  base  of  a  similar  triangle  equal  to  their  sum. 

47.  Calculate  the  area  of  an  isosceles  triangle  whose  equal  sides  are 
24  and  vertex  angle  is  30°;  60°;  135°;  90°. 

48.  Two  similar  fields  contain  together  579  square  feet  and  their 
ratio  of  similitude  is  rV     Calculate  the  area  of  each  field. 

Si  =     T>  m  Si  +£2  =  193 
02      \.2t        01  4*7 

49.  Calculate  the  area  of  a  parallelogram  whose  diagonals  are  75 
and  200,  and  one  of  their  included  angles  30°. 

(a  +  6)  Vab 


50.  Show  that  T 


where  T  is  the  area  of  a  right 


triangle,  and  a  and  b  are  the  segments  of  the  hypotenuse  formed 
by  the  altitude.  Calculate  the  area  of  the  triangle  if  these 
segments  are  4  and  6. 

51.  When  Dido  and  her  followers  landed  on  the  coast  of  Africa, 
"they  asked  of  the  natives  only  so  much  land  as  they  could 
enclose  with  a  bull's  hide"  (Bullfinch,  "Age  of  Fable").  It  is 
told  that  the  hide  was  cut  into  strips.  If  this  hide  contained 
50  square  feet  and  the  strips  were  -fa  inch  wide,  what  area  circle 
was  obtained?  What  area  square  could  be  enclosed? 


AREAS 


289 


52.  A  square  is  2"  on  a  side.     Find  the  side  of  a  second  square 
whose  area  is  1^  times  the  area  of  the  first  square. 

53.  Find  the  side  of  a  square  equivalent  to  the  sum  of  two  squares 
whose  sides  are  3"  and  5"  respectively. 

54.  Calculate  the  altitude  of    a  triangle  whose  base  is  20",  and 
which  is  equivalent  to  a  triangle  whose  altitude  is  12"  and 
base    16". 

55.  Calculate    the    altitude    of    a    triangle    whose    base    is    160" 
and  which  is  equivalent  to  the  sum  of  two  triangles  which 
measure  respectively,  altitude  50",  base  120",  and  altitude  75", 
base  60". 

56.  Find  the  area  of  the  segment  ACB  cut  off  by  a  chord  which 
subtends  an  arc  of  120°,  in  a  circle  whose 

radius  is  20  inches. 

Helps. — Area  of  segment  ACB  =  area  of 
sector  OACB  -  area  of  AOAB.  Alti- 
tude OD  of  AOAB  =  i  radius;  why? 
Find  the  base  of  AOAB;  etc. 

57.  A  stake  is  driven  on  the  farther  edge  of 
a  stream  50  feet  wide,  and  a  horse  on  the 
nearer  side  of  the  stream  is  tied  to  the 

stake  by  a  rope  100  feet  long.     Over  what  area  can  the  horse 
graze  without  crossing  the  stream  ? 

58.  Find  the  area  of  the  minor  segment  cut  off  by  a  chord  which 
subtends  an  arc  of  90°  in  a  circle  of  10  inches  radius. 

59.  What  fractional  parts  of  the  area  of  the  circle 
are  included  between  each  pair  of  equidistant 
parallel  chords  in  this  figure? 

60.  What  area  is  included  between  a  circle  of  10 
inches   radius   and  the  sides  of   an  inscribed 
square? 

61.  Find  the  area  of  an  annulus  whose  radii  are  12  and  20  inches 
respectively? 

62.  Find  the  area  of  a  triangle  whose  sides  are  102,  104,  106. 
Help.— Result  is  S  =  23'32-13'5  =  4680. 

63.  Find  the  side  of  a  square  which  is  equivalent  to  the  sum  of  two 
triangles  whose  bases  and  altitudes  are  respectively  10  and  20 
inches,  and  8  and  20  inches. 

64.  Write  a  formula  expressing  the  area  of  a  parallelogram  of  which 
two  sides  are  a  and  b  and  the  included  angle  is  60°. 

65.  Write  a  formula  expressing  the  area  of  a  parallelogram  of  which 
two  sides  are  a  and  6  and  the  included  angle  is  45°. 

66.  The  perimeter  of  a  rectangle  is  480  feet  and  the  area  is  8000 
square  feet.     Find  the  dimensions  of  the  rectangle. 

19 


290 


PLANE    GEOMETRY 


67.  Two  homologous  sides  of  two  similar  pentagons  are  6  and  10 
inches  respectively.     The  area  of  the  smaller  pentagon  is  720 
square  inches.     What  is  the  area  of  the  larger  pentagon? 

68.  If  the  areas  of  two  similar  quadrilaterals  are  1620  and  288 
square  inches,  respectively,  what  is  the  ratio  of  any  two  homolo- 
gous sides  of  the  quadrilaterals? 

69.  A  line  is  drawn  parallel  to  a  side  of  a  triangle  and  divides  the 
other  two  sides  into  segments  whose  ratio  is  2  to  3.     What  is 
the  ratio  of  the  areas  into  which  the  triangle  is  divided  ?     There 
are  two  positions  of  the  line. 

70.  Where  must  a  line  be  drawn,  parallel  to  one  side  of  a  triangle, 
in  order  to  bisect  the  area  of  the  triangle? 


APPLICATIONS 

382.  Construction  Problems  in  Areas. — These  construc- 
tions are  made  by  the  rule  and  compass  method,  and  are 
chiefly  interesting  because  of  their  association  with  the 
early  development  of  geometry.  Almost  all  problems  in 
combining  and  dividing  areas,  equivalence  of  areas,  etc., 
which  occur  in  practice  are  better  performed  by  geometrical 
or  trigonometrical  calculation.  Such  constructions  as  are 
here  given  may  sometimes  be  valuable  as  checks  on 
calculation. 

1.  To  construct  a  square  equivalent  to  the  sum  of  two  given 
squares. 


A  and  B  are  the  given  squares.     Prove  that  C  =c=  A  +  B. 

2.  To  construct  a  square  equivalent  to  the  difference  of  two 
given  squares. 

Show  how  this  construction  can  be  made. 

3.  To  construct  a  square  equivalent  to  the  sum  of  any 
number  of  given  squares. 

Show  how  the  required  square  is  obtained. 


AREAS 


291 


4.  To  construct  a  triangle  equivalent  to  a  given  triangle  and 
with  a  given  base. 


b' 

Explain  the  construction. 

5.  To  construct  a  triangle  equivalent  to  a  given  rectangle 
and  with  a  given  base. 

Show  how  this  construction  can  be  made. 

6.  To  construct  a  square  equivalent  to  a  given  rectangle. 
Help. — Use  the  principle  of  316,  Exercise  12,  to  obtain 

the  side  of  the  required  square. 

7.  To  construct  a  triangle  equiv- 
alent to  a  given  polygon. 

Show   that  AFCG  =0=   polygon 
ABCDE. 

8.  To  construct  a  square  equiv- 
alent to  a  given  triangle. 

Helps. — Let  a  and  b  be  the  altitude  and  base  of  the  given 
triangle,  and  s  be  the  side  of  the  required  square. 


3 


F  A 


Then 


ab  ,  a        s 

~2   =  8  '  and  ~8  =  T" 


Use  the  method  of  Problem  6  to  complete  the  construction. 

EXERCISES 

1.  Construct  a  square  equivalent  to  the  sum  of  two  given  triangles. 

2.  Construct  a  circle  equivalent  to  the  sum  of  two  given  circles. 

3.  Construct  an  equilateral  triangle  equivalent  to  a  given  triangle. 

4.  Bisect  a  triangle  by  a  line  parallel  to  a  side. 

5.  Bisect  a  quadrilateral  by  a  line  drawn  from  a  given  vertex. 


292 


PLANE    GEOMETRY 


6.  Construct  an  isosceles  triangle  of  which  the  vertex  angle  equals 
an  angle  of  a  given  triangle,  and  which  is  equivalent  to  the  given 
triangle. 

Helps.— (1)  c2  =  ab;  (2)  to  find  c,  refer  to  316,  Exercise  12,  or 
Exercise  13. 


383.  Surveying. — When  the  shape  of  a  piece  of  land  is 
rectangular,  triangular,  circular,  etc.,  its  area  is  found  by 
methods  already  given  in  this  chapter.  The  shape  of  a 
parcel  of  land  is  generally  irregular,  and  special  methods 
have  been  devised  for  computing  the  areas  of  such  plots 
from  linear  and  angular  measurements. 

1.  Areas  Bounded  by  Straight  Lines,  (a)  Area  by 
Coordinates. — Take  vertex  A,  as  the  origin  of  coordinates. 
By  measuring  the  sides  and 
interior  angles  of  the  field 
ABCDE,  it  is  possible  to  calcu- 
late by  trigonometry  the  coor- 
dinates of  all  the  vertices.  The 
lines  B'B,  C'C,  D'D,  E'E,  and 
AB',  B'C',  C'D',  D'E',  E'A,  can 
then  be  found.  These  lines  are 
the  dimension  lines  of  the  part 
areas,  ABB',  B'BCC',  C'CDD', 
D'DEE',  E'EA,  and  from  them  the  values  of  these  areas 
can  be  found.  The  polygon  ABCDE  is  an  algebriac  sum 
of  the  part  areas. 

EXERCISES 

1.  The  coordinates  (abscissa  and  ordinate)  of  the  vertices  of  a 
field  ABCDE,  are  A,  (0,  0);  B,  (  +  100,  +80);  C,  (  +  150,  +90); 
D,  (  +  175,  +20);  #,(  +  120,  -50).     Plot  the  vertices  on  squared 
paper  and  draw  the  field. 

2.  Find  the  values  of  the  lines  B'B,  C'C,  D'D  and  E'E. 

3.  Find  the  values  of  the  lines  AB',  B'C',  C'D',  D'E'  and  E'A. 

4.  Find  the  areas  ABB',  B'BCC',  C'CDD',  D'DEE'  and  E'EA. 


AREAS 


293 


5.  Find  the  area  of  the  field  ABCDE  from  the  algebraic  sum  of  the 
part  areas. 

Help.—  Polygon  A  BCDE  =  -  AABB'  -  trapezoid  B'BCC'  + 
trapezoid  C'CDD',  etc. 

(6)  Area  by  Triangles.  —  By  this  method  the  field  is 
divided  into  triangles,  (a)  by  diagonals,  (6)  by  lines  drawn 
to  every  vertex  from  some  interior  point,  (c)  in  any 
convenient  way. 

EXERCISES 


6.  The   measurements 
ZBOC  =  60° 


of   a   field    ABCDE 
=  90°,   /.DOE  = 


are     Z.AOB  =  60 


60°,    /.EOA  =  90°;  OA  =  120  feet,  OB 
=   120  feet,  OC  =  80  feet,  OD  =  150 
feet,  OE  =  1  00  feet.     Find  the  area  by 
finding  the  sum  of  the  areas  of  the  tri- 
angles.    Trigonometry  is  required  for 
this  calculation  when  the  central  angles 
are  other  than  30°,  45°,  60°,  90°,  and  their  supplements. 
7.  Find  the  area  of  a  four-sided  field  A  BCD,  of  which  the  measure- 
ments are  AB  =  400  feet,  BC  =  1500  feet,  CD  =  1200  feet, 
DA  =  800  feet,  and  diagonal  BD  =  1000  feet.     Find  the  areas 
of  triangles  ABD  and  BCD  by  the  formula  of  367. 

2.  Areas  Bounded  by  Irregu- 
lar Curved  Lines.  —  The  methods 
used  give  the  area  approxi- 
mately, but  with  sufficient  exact- 
ness. They  apply  to  plots  of 
either  of  these  forms. 

(a)  The  Ordinate  Method  — 
Measurements  are  taken  in  the 
field,  AD,  DF,  FH,  etc.,  and  per- 
pendiculars  AC,  DE,  etc.  The 
field  is  then  plotted  accurately  to 
scale,  and  the  line  AB  is  divided 
into  equal  parts  and  the  ordinates 
Oi,  02,  etc.,  are  drawn  at  the  mid- 
dle points  of  the  equal  segments. 
The  area  equals  the  average  ordi- 
nate  multiplied  by  the  uniform  segment  x. 


FIG.  1. 


FIG.  2. 


294  PLANE   GEOMETRY 

EXERCISES 

8.  The  measurements  of  a  field  (Fig.  1)  are  AD  =  DF  =  FH  = 
etc.  =  40  feet;  ordinates  ylt  y2,  etc.,  are  80,  90,  120,  140,  140, 
110,  75,  70,  30  feet.     Plot  the  field  on  squared  paper. 

9.  Find  the  area  of  the  field  of  Exercise  8  by  the  ordinate  method. 

(6)  The   Trapezoid   Method.  —  Each   part   area,   AC  ED, 
DEGF,  etc.,  is  approximately  a  trapezoid. 

EXERCISES 

10.  Show  that  the  area  of  a  field  of  either  form,  when  AD,  DF,  etc. 
are  all  equal  to  x,  is  given  by  the  formula, 

S   =   \  h/l+2/n  +2(7/2  +2/3+     ....    •</«-!)]• 

11.  Find  the  area  of  the  field  of  Exercise  8  by  the  trapezoid  method. 
(c)  Simpson's    Rule.  —  This    is     considered     the     most 

accurate  method.     The  formula  for  the  area  is, 

S  =       [y\  +  yn  +  4(i/2  +  2/4  +  etc.)  +  2(y3  +  yb  +  etc.)] 


=     [A  +  4B  +  2C]- 


EXERCISES 

12.  For  what  do  the  letters  A,  B,  C,  stand  in  the  final  formula? 

13.  Calculate  the  area  of  the  field  of  Exercise  8  by  Simpson's  Rule. 

3.  'Special  Methods  of  Finding  Areas,  (a)  Area  by 
Squared  Paper.  —  The  field  is  plotted  and  the  area  measure 
is  counted. 

EXERCISE 

14.  Find  the  area  of  the  field  of  Exercise  8  by  this  method. 

(b)  Area  by  Planimeter.  —  An  area  may  be  measured 
automatically  by  this  instrument,  by  tracing  the  bounding 


d 

line  with  the  point  A  while  the  point  B  is  fixed.     The  re- 
volving counter  C  records  the  area.     Explanations  of  the 


AREAS  295 

planimeter    may    be    found    in    texts    on    surveying    and 
engineering. 

(c)  Area  by  Approximate  Rectangle,  Triangle,  Circle,  etc. — 
The  area  is  plotted  accurately,  and  straight  lines  are  drawn 
which  approximate  as  closely 
as  possible  to  the  bounding 
curve,  while  the  form  of  the 
rectilinear  figure  is  made  a 
rectangle,  triangle. or  trape- 
zoid;  or  in  some  cases  an  approximate  circle  or  semi-circle 
may  be  used.  The  dimensions  of  this  approximate  figure 
are  scaled  from  the  drawing. 

EXERCISE 

15.  Find  the  area  of  the  field  of  Exercise  8  by  drawing  an  approxi- 
mate rectangle. 

384.  Dividing  Land. — Problems  of  dividing  land  occur 
constantly  in  surveying.  They  are  solved  by  methods  of  cal- 
culation, and  may  be  checked  by  geometrical  constructions. 

EXERCISES 

1.  Divide  a  rectangular  field  120  feet  by  240  feet  into  three  equiva- 
lent parts  by  lines  parallel  to  the  shorter  side. 

2.  Divide  a  rectangular  field  120  feet  by  240 
feet,  into  three  equivalent  parts  by  lines 
drawn  from  a  vertex  of  the  field.     Find 
BF  and  DE.  ^ 

3.  The  sides  of  a  triangular  field  are  AB  = 

150  feet,  BC  =  200  feet,  CA  =  210  feet.     Divide  it  into  two 
equivalent  parts  by  a  line  drawn  from  the  vertex  A. 

4.  Divide  the  field  of  Exercise  3  into  two  equivalent  parts  by  a 
line  DE  parallel  to  BC.     Calculate  AD  and  AE. 

5.  Divide  the  field  of  Exercises  3  and  4  into  two  equivalent  parts 
by  a  line  drawn  from  a  point  F,  which  is  20  feet  from  B  on  the 
side  AB. 

385.  Estimating  and  Engineering. 

EXERCISES 

1.  A  room  is  30  feet  long,  20  feet  wide,  10  feet  high.     Calculate 
the  areas  of  the  four  walls  and  ceiling.     There  are  4  windows 


296 


PLANE    GEOMETRY 


3|  feet  by  6  feet,  and  2  doors  4  feet  by  1\  feet.  What  area  is 
covered  by  plaster?  Estimate  its  cost  at  95  cents  per  square 
yard. 

2.  A  barn  is  40  feet  wide,  70  feet  long,  and  24  feet  high  to  the 
eaves;  the  rise  of  the  roof  is  12  feet. 

Estimate  the  cost  of  painting  the 
barn  two  coats  if  1  gallon  of  paint 
will  cover  400  square  feet  one  coat, 
and  paint  costs  $2.75  per  gallon. 

3.  Shingles  average  6  by  24  inches.     When  laid  on  a  roof,  \  of  the 
shingle  is  exposed.     Estimate  the  cost  of  shingling  the  roof  of 
the  barn  of  Exercise  2  at  $20.00  per  1000  shingles. 

4.  A  rectangular  piece  of  sheet  metal  4  inches  by  5  inches,  weighs 
\  a  pound.     What  is  the  weight  of  a  rectangular  piece  of  the 
same  sheet  metal  10  inches  by  12  inches? 

5.  Find  the  number  of  square  yards  covered  by  a  rectangular 
building  120  feet  by  55  feet.     Each  yard  in  depth  of  cellar  ex- 
cavation forms  a  cubic  yard  with  a  square  yard  of  surface. 
Estimate  the  number  of  cubic  yards  excavated  in  digging  the 
cellar  6  feet  deep;  and  the  cost  at  75  cents  per  cubic  yard. 

6.  Make  a  similar  cost  estimate  for  the  excavation  of  a  cellar  6  feet 

SO 


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38 

deep  for  the  building  of  which  the  floor  plan  is  here  given. 

7.  The  number  of  gallons  of  water  flowing  past  a  given  line  across 
a  river  is  found  by  measuring  its  "cross-section."     Find  the 
cross-section  of  a  river,    the    soundings  being  taken  from  a 
bridge  at  intervals  of  50  feet  from  shore  to  shore.     Depths  in 
feet:  10,  20,  25,  30,  20,  18,  15,  10,  5.     Total  width  500  feet. 
If  the  velocity  of  current  is  200  feet  per-  minute,  how  many 
cubic  feet  of  water  flow  past  the  bridge  per  minute? 

8.  If  10  per  cent  of  the  water  of  Exercise  7  can  be  pumped  from 
the  river  to  supply  a  city,  is  there  sufficient  water  for  a  city  of 
1,000,000  inhabitants,  allowing  200  gallons  per  person  per  day 
of  24  hours? 


AREAS 


297 


9.  A  playground  is  trapezoidal  in  shape,  the  parallel  sides  being 
300  feet  and  240  feet,  and  a  third  side  perpendicular  to  them 
250  feet.  How  many  children  can  be  accommodated  if  200 
square  feet  is  the  estimated  allotment  per  child? 


S',5 

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5 

ac 

10.  A  cement  walk  4  ft.  wide  is  laid  along  two  sides  of  a  corner  lot. 
Calculate  the  cost  at  25  cents  per  square  foot. 

11.  A  cement  walk  5  feet  wide  runs  straight  for  80  ft.,  then  follows 
a  circular  curve  of  an  inner  radius  of  50  feet  for  a  quarter  circle, 
then  runs  straight  for  100  feet.     Estimate  the  cost  at  20  cents 
per  square  foot. 

12.  Parquet  floors  of  white  oak  are  to  be  laid  in  rooms  as  follows: 
two  rooms  30  by  12  feet;  three  rooms  22  by  18  feet;  one  room 
35  by  15  feet;  one  hall  6  by  24  feet.     Estimate  cost  if  flooring 
2  inches  wide  costs  $25  per  1000  linear  feet,  and  an  additional 
charge  is  made  of  10  cents  per  square  foot  for  laying  and 
finishing. 

13.  A  circular  rod  in  a  roof  truss  must  sustain  a  safe  tension  of 
20,000  pounds  per  square  inch  of  cross-section.     What  must 
be  its  diameter  if  the  total  tension  in  the  rod  is  60,000  pounds? 
What  if  the  tension  is  12,000  pounds? 

14.  How  many  boiler  tubes  3  inches  in  diameter  are  required  to  give 
a  total  cross-section  equal  to  that  of  a  3-foot  diameter  chimney? 

15.  Three  pipes,  10  inches,  12  inches,  8  inches,  diameter,  respec- 
tively, empty  into  a  basin.     What  diameter  must  be  given  to 
a  single  discharging  pipe  whose  cross-sectional  area  equals  the 
combined  areas  of  the  three  pipes? 

16.  What  total  pressure  is  exerted  by  the  steam  in  the  cylinder  of  a 
steam  engine  against  a  piston  12  inches  in  diameter  if  the  pres- 
sure is  120  pounds  per  square  inch?     The  horsepower  of  the 

in  which  p  is  the 

average  steam  pressure,  I  the  length  of  stroke,  a  the  area  of  the 
piston,  n  the  number  of  strokes  per  minute.  Find  the  horse- 
power of  the  engine  considered  if  the  length  of  stroke  is  20 
inches  and  number  of  strokes  is  160  (number  of  revolutions  80) 
per  minute. 

17.  A  transit  telescope  magnifies  7  diameters.     How  many  times 
is  the  area  magnified? 


engine  is  given  by  the  formula  h.p.  =  oo  QQQ 


298  PLANE    GEOMETRY 

18.  Show  that  the  formula  for  the  area  of  an  annulus  may  be  ex- 
pressed, A  =  TT-—      —'t;  where  di  and  dz  are  the  outer  and  inner 

& 

diameters  and  t  is  the  thickness  (width  of  ring).     This  is  often 
used  as  a  shop  formula. 

19.  The  cross-section  area  of  an  I-beam  is  required  in  order  to  find 


T 

<-<*     ^ 


the  weight  per  yard  of  length.     Show  that  the  area  of  cross- 
section  is  given  by  the  formula  A  =  de  +  2c(a  +  6). 
20.  Find  the  cross-section  area  of  an  I-beam  if  a  =  I",  b  =  1|", 
c  =  li",  d  =  *",  e  =  10". 

PROBLEMS  FOR  FIELD  WORK 

386.  (1)  Find  the  area  of  a  plot  of  ground  bounded  by 
straight  lines,  by  measuring  all  of  the  sides  of  the  plot  and 
as  many  diagonals  as  are  necessary  to  divide  it  into  tri- 
angles. The  diagonals  should  be  selected  so  as  to  form  as 
nearly  as  possible  equilateral  triangles.  Use  the  formula 
of  367. 

2.  Find  the  area  of  a  piece  of  land  entirely  bounded  by 
a  curved  line,  or  by  straight  and  curved  lines.  Use  the 
method  of  383,  Part  2(6). 


INDEX  OF  PRINCIPAL  FORMULAS 

Arc 

length  of,  330,  331 

radius  of,  315,  Exercise  42,  322 

rise  of,  322 
Areas 

of  an  annulus,  372,  Theorem  (14) 

of  a  circle,  370 

of  a  circumscribed  polygon,  369 

of  an  equilateral  triangle,  372,  Theorem  (10) 

of  a  figure  bounded  by  straight  and  curved  lines,  383 

of  a  parallelogram,  364 

of  a  polygon,  383 

of  a  rectangle,  352,  363 

of  a  regular  hexagon,  372,  Theorem  (12) 

of  a  regular  polygon,  368 

of  a  segment  of  a  circle,  372,  Theorem  (17) 

of  a  square  in  terms  of  the  diagonal,  372,  Theorem  (7) 

of  a  trapezoid,  366 

of  a  triangle,  365,  367 

of  a  triangle  in  terms  of  the  sides  and  the  radius  of  the  circum- 
scribed circle,  372,  Theorem  (15) 

Chord,  length  of,  322 
Circle 

angle  formed  by  a  tangent  and  a  chord,  297 

angle  formed  by  two  chords,  298 

angle  formed  by  a  secant  and  a  tangent,  300 

angle  formed  by  two  secants,  299,  302,  Theorem  (8) 

angle  formed  by  two  tangents,  301 

area  of,  370 

inscribed  angle,  296,  302,  Theorem  (7) 

length  of  an  arc,  330,  331 

length  of  a  chord,  322 

relation  between  a  half  chord  and  the  segments  of  a  diameter 
perpendicular  to  the  chord,  260,  Theorem  (20),  305 

relation  between  the  segments  of  a  secant  and  an  intersecting 
tangent,  307 

relation  between  the  segments  of  two  intersecting  chords,  305 

relation  between  the  segments  of  two  intersecting  secants,  306 

299 


300  INDEX    OF    PRINCIPAL    FORMULAS 

Circle 

relations  between  radius,  diameter  and  circumference,  337 

value  of  TT,  340 
Curvature  of  a  sphere,  317 
Curves 

compound  curves,  282,  Exercise  9 

laying  out  a  circular  arc,  282 

laying  out  a  circular  arc  by  using  two  transits,  282 

laying  out  a  circular  arc  by  ordinates,  319 

laying  out  a  circular  arc  by  offsets  from  the  tangent,  322 

7-curves,  282,  Exercise  12 

Horizon  at  sea,  distance  of,  317 

Polygon 

area  of,  383 

sum  of  the  interior  angles  of,  187 

sum  of  the  exterior  angles  of,  188 
Proportion,  laws  of,  240 
Pythagorean 

numbers,  261,  Exercise  14 

theorem,  250 

Sect 

division  into  mean  and  extreme  ratio,  316 
harmonic  division,  260,  Theorem  (33) 
projection  of,  254,  265,  Exercises  22,  23,  24 

Square 

diagonal  in  terms  of  a  side,  260,  Theorem  (14),  265,  Exercise  28 

side  in  terms  of  the  diagonal,  260,  Theorem  (17) 

side  of  an  octagon  obtained  from,  261,  Exercises  61,  62 

Trapezoid 

area  of,  366 

relation  between  lengths  of  median  and  bases,  184 
Triangle 

altitude,  length  of,  255 

altitude   of   an   equilateral   triangle,   260,    Theorem    (16),    265, 
Exercise  31 

area  of,  365,  367 

bisector  of  an  interior  angle,  length  of,  309,  310 

circumscribed  circle,  radius  of,  372,  Theorem  (16) 

circumscribed  circle  of  an  equilateral  triangle,  radius  of,  315, 
Exercise  24 

inscribed  circle,  radius  of,  371 


INDEX    OF   PRINCIPAL    FORMULAS  301 

Triangle 

median,  length  of,  256 

medians,  point  of  concurrence  of,  185 

segments  into  which  a  bisector  of  an  interior  angle  divides  the 
opposite  side,  245,  260,  Theorem  (26) 

segments  into  which  a  bisector  of  an  exterior  angle  divides  the 
opposite  side,  246 

side  of  an  equilateral  triangle  expressed  in  terms  of  the  altitude, 
260,  Theorem  (18) 

side  opposite  an  acute  angle,  length  of,  252,  265,  Exercise  25 

side  opposite  an  obtuse  angle,  length  of,  253 

sides  of  a  right  triangle,  relation  between,  250 

solution  of,  by  trigonometry,  265 
Trigonometry 

distance  across  a  pond,  etc.,  265,  Exercise  17 

distance  across  a  river,  etc.,  265,  Exercise  4 

distance  to  an  island,  etc.,  265,  Exercise  18 

height  of  an  aeroplane,  etc.,  265,  Exercise  7 

height  of  a  pole,  etc.,  265,  Exercise  5 

height  of  a  stack,  etc.,  265,  Exercise  20 

rise  of  a  hill,  265,  Exercise  6 

solution  of  an  oblique  triangle  when  two  sides  and  the  included 

angle  are  given,  265,  Exercises  25,  27 
Trisection  of  an  angle,  316 


, 


1C  49582 


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